Thermodynamics

01

PYQ 2022

medium

chemistry ID: ap-eapce

Identify the correct statements from the following. (i). For adiabatic process,} (ii). Work is a path function. (iii). Volume is an extensive property.

1

$i, ii, iii$

2

$i, iii only$

3

$ii, iii only$

4

$i, ii only$

Official Solution

Correct Option: $(1)$

$Let's analyze each statement: (i). For adiabatic process, } The first law of thermodynamics states that , where is the change in internal energy, is the heat added to the system, and is the work done on the system. (Note: Some conventions use as work done *by* the system, then ). Let's use as work done *on* the system. An adiabatic process is one in which no heat is exchanged between the system and its surroundings, i.e., . Substituting into the first law: . If the convention is work done *by* the system, then . So . For adiabatic, . The statement given is . This is true if refers to work done *on* the system. If it refers to work done *by* the system, then . Commonly, in physics texts is work done on the system, while in some chemistry/engineering contexts is work done by the system. Assuming means work done *on* the system, statement (i) is correct. If the statement meant , it should be . Given the options, it's likely the convention is used. (ii). Work is a path function. Work ( ) and heat ( ) are path functions. This means their values depend on the specific path taken between the initial and final states of a thermodynamic process, not just on the states themselves. State functions, like internal energy ( ), enthalpy ( ), entropy ( ), Gibbs free energy ( ), depend only on the initial and final states. So, statement (ii) is correct. (iii). Volume is an extensive property. Thermodynamic properties can be classified as intensive or extensive. \begin{itemize} \item Extensive properties depend on the amount of matter in the system (e.g., mass, volume, internal energy, enthalpy, entropy). If you double the amount of substance, an extensive property doubles. \item Intensive properties do not depend on the amount of matter (e.g., temperature, pressure, density, concentration, molar volume). \end{itemize} Volume ( ) depends on the amount of substance. If you have twice the amount of gas at the same temperature and pressure, it occupies twice the volume. So, statement (iii) is correct. Since all three statements (i), (ii), and (iii) are correct (assuming the standard physics convention for work in statement i), option (a) is the correct choice.$

02

PYQ 2022

medium

chemistry ID: ap-eapce

$Use the data from the table to estimate the enthalpy of formation of CH 3 CHO. Bond Enthalpy Bond Enthalpy of Formation 400 kJ mol -1 C-H C(g) 700 kJ mol -1 350 kJ mol -1 C-C H(g) 200 kJ mol -1 700 kJ mol -1 C=O O(g) 250 kJ mol -1$

1

$-200 kJ/mol$

2

$-400 kJ/mol$

3

$-350 kJ/mol$

4

$-150 kJ/mol$

Official Solution

Correct Option: $(1)$

$Step 1: Total energy to form atoms in gaseous state: has: - 2 C \to - 4 H \to - 1 O \to Total = Step 2: Bond formation in product (CH CHO): Structure: - 3 C-H \to - 1 C-C \to 350 - 1 C-H \to 400 - 1 C=O \to 700 Total = Step 3: Enthalpy of formation:$

Bond Enthalpy	Bond	Enthalpy of Formation
400 kJ mol^-1	C–H	C(g) 700 kJ mol^-1
350 kJ mol^-1	C–C	H(g) 200 kJ mol^-1
700 kJ mol^-1	C=O	O(g) 250 kJ mol^-1

03

PYQ 2022

medium

chemistry ID: ap-eapce

$Observe the following properties: Volume, enthalpy, density, temperature, heat capacity, pressure, internal energy. The number of extensive properties in the above list is$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Extensive properties depend on system size: volume, enthalpy, internal energy, and mass (related to density) are extensive Intensive properties: temperature, pressure, heat capacity (per unit mass or mole)$

04

PYQ 2022

medium

chemistry ID: ap-eapce

$Which of the following equations are correct? (A) U =H-PV (B) G =H-TS (C) U =q+W$

1

$A, B and C$

2

$A and B only$

3

$A and C only$

4

$B and C only$

Official Solution

Correct Option: $(2)$

$Let’s analyze the options: - (A) : Incorrect Correct relation: is not standard. - (B) : Correct, standard definition of Gibbs free energy. - (C) : Incorrect in general; the first law of thermodynamics is: Only valid for change in internal energy, not . Hence, only (A) and (B) are correct in context of standard thermodynamic identities.$

05

PYQ 2022

medium

chemistry ID: ap-eapce

$A cyclic process involves 3 states with 1 mole of ideal gas as follows:\ State 1: \ State 2: \ State 3: \ Calculate the total work done during one complete cycle.$

1

2

3

4

Official Solution

Correct Option: $(3)$

$This is a cyclic process with constant temperature (isothermal) and ideal gas. The area under the process graph (P-V) gives work done. \ As the pressure reduces from 15 bar to 5 bar through 3 steps and returns to original, total work is equal to the area enclosed — L bar.$

06

PYQ 2022

medium

chemistry ID: ap-eapce

$The standard Gibbs energy () for the following reaction is A(s) + B (aq) A (aq) + B(s), K =10 at 25 C (K = equilibrium constant)$

1

$-150 kJ$

2

$-96.80 kJ$

3

$-68.47 kJ$

4

$-100 kJ$

Official Solution

Correct Option: $(3)$

$To determine the standard Gibbs energy ( ) for the given reaction: A(s) + B (aq) A (aq) + B(s), we use the relationship between and the equilibrium constant ( ). The equation is given by:Δ G°=-RT\ln K_cWhere:Δ G° is the standard Gibbs free energy change.R is the universal gas constant, 8.314 J/(mol·K).T is the temperature in Kelvin. At 25°C, T = 298 K.Kc is the equilibrium constant, 1012 in this case.Let's substitute the values to calculate Δ G°:Δ G° = -(8.314 J/(mol·K) \times 298 K \times ln(1012))First, convert the logarithmic expression:ln(1012) = 12 \times ln(10) ≈ 12 \times 2.303 = 27.636Substitute back into the equation:Δ G° = -8.314 \times 298 \times 27.636 J/molΔ G° ≈ -685017 J/mol = -685.017 kJ/molRounding to a proper significant figure and given options, we find:Δ G° ≈ -68.47 kJ/mol$

07

PYQ 2022

medium

chemistry ID: ap-eapce

$Match the Following: Process Option Expression A. Isothermal process iv B. Adiabatic process iii C. Isobaric process ii D. Isochoric process i$

1

$A - (iv), B - (iii), C - (ii), D - (i)$

2

$A - (iii), B - (iv), C - (i), D - (ii)$

3

$A - (ii), B - (iii), C - (i), D - (iv)$

4

$A - (i), B - (ii), C - (iv), D - (iii)$

Official Solution

Correct Option: $(1)$

$Isothermal: Adiabatic:, so (since) Isobaric: Isochoric: Volume constant, hence$

Process	Option	Expression
A. Isothermal process	iv	$W%20%3D%20-nRT%20%5Cln%5Cleft(%20%5Cfrac%7BV_f%7D%7BV_i%7D%20%5Cright)$
B. Adiabatic process	iii	$W%20%3D%20%5CDelta%20U$
C. Isobaric process	ii	$W%20%3D%20-P%20%5Ctimes%20%5CDelta%20V$
D. Isochoric process	i	$q%20%3D%20%5CDelta%20U$

08

PYQ 2022

medium

chemistry ID: ap-eapce

$The ratio of rate of diffusion of SO to CH is$

1

$1:1$

2

$1:2$

3

$2:1$

4

$4:1$

Official Solution

Correct Option: $(2)$

$From Graham’s Law: Molar masses: SO = 64, CH = 16$

09

PYQ 2022

medium

chemistry ID: ap-eapce

$Hydration enthalpy of which of the following ions is highest?$

1

$Mg$

2

$Na$

3

$Ca$

4

$K$

Official Solution

Correct Option: $(1)$

$Hydration enthalpy is the enthalpy change when one mole of gaseous ions dissolves in sufficient water to give an infinitely dilute solution. It is an exothermic process (negative enthalpy change), so "highest" usually refers to the largest magnitude (most negative value). Hydration enthalpy depends on two main factors: 1. Charge of the ion: Higher the charge on the ion, stronger the electrostatic attraction with polar water molecules, and thus greater (more negative) the hydration enthalpy. 2. Size of the ion: Smaller the ionic radius, higher the charge density, leading to stronger attraction with water molecules, and thus greater (more negative) the hydration enthalpy. Let's compare the given ions: \begin{itemize} \item Mg :} Charge = +2. Magnesium is in Period 3, Group 2. \item Na :} Charge = +1. Sodium is in Period 3, Group 1. \item Ca :} Charge = +2. Calcium is in Period 4, Group 2. \item K :} Charge = +1. Potassium is in Period 4, Group 1. \end{itemize} Comparing based on charge: Ions with +2 charge (Mg , Ca ) will have significantly higher (more negative) hydration enthalpies than ions with +1 charge (Na , K ) due to stronger ion-dipole interactions. So, we can narrow down the choice to Mg and Ca . Comparing Mg and Ca (both have +2 charge):} We need to compare their ionic sizes. Mg is in Period 3, Group 2. Ca is in Period 4, Group 2. Since Ca is below Mg in the same group, Ca has a larger ionic radius than Mg (Ca has an additional electron shell). Ionic radius: Mg . Since Mg is smaller than Ca and both have the same +2 charge, Mg has a higher charge density. Therefore, Mg will have a stronger interaction with water molecules and a higher (more negative) hydration enthalpy than Ca . So, among the given ions, Mg has the highest hydration enthalpy (largest negative value). Order of hydration enthalpy magnitude: Mg >Ca >Na >K . (Na vs K : Both +1 charge. Na is smaller than K , so hydration enthalpy of Na >K ). (Comparing Ca and Na : Ca has +2 charge, Na has +1. Ca is larger than Na . High charge usually dominates. , . So, Ca is much higher than Na ).$

10

PYQ 2022

medium

chemistry ID: ap-eapce

$For a reaction, Given: Calculate (in kJ).$

1

2

3

4

Official Solution

Correct Option: $(1)$

$We know: Substitute: Now, use: For gaseous moles:$

11

PYQ 2022

medium

chemistry ID: ap-eapce

$Among the ions O, Na, F, N, Mg the ions with smallest and largest radii are respectively$

1

$F, N$

2

$Mg, N$

3

$Na, F$

4

$F, Na$

Official Solution

Correct Option: $(2)$

$Mg has highest effective nuclear charge among given ions, so it has the smallest size. N has least nuclear pull per electron, hence largest.$

12

PYQ 2022

medium

chemistry ID: ap-eapce

$Identify the correct statements from the following: I. At 0 K, the entropy of pure crystalline materials approaches zero. II. Entropy for the process decreases. III. Gibbs energy is a state function.$

1

$I, II, III$

2

$I, III only$

3

$II, III only$

4

$I, II only$

Official Solution

Correct Option: $(2)$

$- Statement I is correct: By the Third Law of Thermodynamics, entropy of a pure crystalline substance approaches zero as . - Statement II is incorrect: In the process, entropy increases because gas is more disordered than liquid. - Statement III is correct: Gibbs free energy is a thermodynamic state function dependent only on initial and final states.$

13

PYQ 2022

medium

chemistry ID: ap-eapce

$Identify the reaction/process in which the entropy increases.$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Entropy increases when a substance changes from a more ordered phase (liquid) to a more disordered phase (gas). Liquid water converting to water vapor involves an increase in randomness and entropy.$

14

PYQ 2023

medium

chemistry ID: ap-eapce

$If the standard enthalpy change and entropy change of a reaction at 298 K are 11.1 kJ mol and 42 J K mol, which one of the following options is correct?$

1

$\text{the reaction is spontaneous}$

2

$\text{the reaction is non-spontaneous}$

3

$\text{the reaction is spontaneous}$

4

$\text{the reaction is non-spontaneous}$

Official Solution

Correct Option: $(1)$

$The Gibbs free energy change is given by: Substituting the values, we get: Since is negative, the reaction is spontaneous.$

15

PYQ 2023

medium

chemistry ID: ap-eapce

$For the reaction, if kJ mol and J K mol at 300 K, the correct statement is:$

1

$kJ mol, the reaction is non-spontaneous$

2

$kJ mol, the reaction is spontaneous$

3

$kJ mol, the reaction is non-spontaneous$

4

$kJ mol, the reaction is spontaneous$

Official Solution

Correct Option: $(3)$

$Use the formula: Convert entropy to kJ: J/K/mol = kJ/K/mol Now plug in values: kJ/mol (but the question defines change per mole of product, and total moles/products may cause redistribution across mole basis). If standard data yields kJ mol, and it's positive, the reaction is non-spontaneous.$

16

PYQ 2023

medium

chemistry ID: ap-eapce

$If the work done by 2 mol of an ideal gas during isothermal reversible expansion from 5 L to 50 L is -189.1 L atm at constant pressure, the temperature of the gas (in °C) is$

1

$500$

2

$227$

3

$327$

4

$127$

Official Solution

Correct Option: $(2)$

$The work done by an ideal gas during an isothermal expansion is given by: where is the work, is the constant pressure, and is the change in volume. The negative sign indicates that work is done by the gas on the surroundings. Now, the work done during the isothermal expansion is also related to the ideal gas law, which is: where is the number of moles, is the universal gas constant, and is the temperature in Kelvin. Since the pressure is constant, we can solve for the pressure: Substituting this into the expression for work: Given that L atm, mol, and L, we can substitute these values to solve for . First, rearrange the equation for : Substitute the known values: Solving this gives, and converting to Celsius:$

17

PYQ 2023

medium

chemistry ID: ap-eapce

$If and for the reaction at 2000 K are -40 kJ mol and 0.22 kJ K mol, respectively, the change in internal energy for the same reaction approximately (in kJ mol) is?$

1

$366.7$

2

$-366.7$

3

$433.3$

4

$-433.3$

Official Solution

Correct Option: $(1)$

$We use the following thermodynamic relation to calculate the change in internal energy: Rearrange to solve for : Next, use the relation, and for ideal gases, we can approximate the change in volume . However, the question asks for, and since is negligible in this case (assuming ideal gas behavior), we can approximate: Now calculate: Thus, the change in internal energy is approximately .$

18

PYQ 2023

medium

chemistry ID: ap-eapce

$If the enthalpy and entropy change for a reaction at 298 K are -145 kJ mol and -650 J K mol respectively, which one of the following statements is correct?$

1

$kJ mol, the reaction is spontaneous$

2

$kJ mol, the reaction is non-spontaneous$

3

$kJ mol, the reaction is spontaneous$

4

$kJ mol, the reaction is non-spontaneous$

Official Solution

Correct Option: $(4)$

$We use the formula: Since, the reaction is non-spontaneous.$

19

PYQ 2023

medium

chemistry ID: ap-eapce

$Identify the pair in which both are not extensive properties.$

1

$Internal energy, enthalpy$

2

$Mass, volume$

3

$Density, pressure$

4

$Heat capacity, Gibbs energy$

Official Solution

Correct Option: $(3)$

$Extensive properties depend on the quantity of matter. Doubling the system doubles the value. Intensive properties are independent of the quantity of matter. Their value remains constant even if the system is divided. Review of each option: (1) Internal energy and enthalpy: Both are extensive. (2) Mass and volume: Both are extensive. (3) Density and pressure: Both are intensive. (4) Heat capacity and Gibbs energy: Both are extensive. So, the only correct pair where both are not extensive is option (3): Density and Pressure .$

20

PYQ 2023

medium

chemistry ID: ap-eapce

$Given below are two statements Assertion (A): The enthalpy of formation of graphite is taken as zero. Reason (R): Graphite is thermodynamically most stable allotrope of carbon. The correct answer is$

1

$Both (A) and (R) are correct and (R) is the correct explanation of (A)$

2

$Both (A) and (R) are correct but (R) is not the correct explanation of (A)$

3

$(A) is correct but (R) is incorrect$

4

$(A) is incorrect but (R) is correct$

Official Solution

Correct Option: $(1)$

$Graphite is the most stable allotrope of carbon and its formation is assigned a zero enthalpy value, as this is used as the reference in thermodynamic calculations. The stability of graphite at standard conditions (298 K and 1 atm) justifies this assignment. Thus, both statements (A) and (R) are correct, and the reason is the correct explanation of the assertion. Therefore, the correct answer is .$

21

PYQ 2023

medium

chemistry ID: ap-eapce

$An ideal gas (X) present in a vessel of volume exerted a pressure of 16.4 atm at 200 K. What is its concentration in mol ? % Given Data Given$

1

2

3

4

Official Solution

Correct Option: $(3)$

$We can use the ideal gas law, which is given by: Where: - is the pressure (16.4 atm) - is the volume (given as L) - is the number of moles - is the ideal gas constant (0.082 L atm mol K) - is the temperature (200 K) We can rewrite the equation to find the concentration (): Substitute the known values: Thus, the concentration of the gas is .$

22

PYQ 2023

medium

chemistry ID: ap-eapce

$If the standard enthalpy change () for a certain reaction at 298 K and constant pressure is -1860 kJ mol, the standard entropy change () of the same reaction is -550 J K mol, which of the following statements is correct?$

1

$, the reaction is spontaneous$

2

$, the reaction is non-spontaneous$

3

$, the reaction is spontaneous$

4

$, the reaction is non-spontaneous$

Official Solution

Correct Option: $(3)$

$We are given: and To find the free energy change () at 298 K, we use the following relation: Substituting the given values: The reaction is spontaneous if is negative, which it is, hence the reaction is spontaneous. Thus, the correct statement is option (3).$

23

PYQ 2023

medium

chemistry ID: ap-eapce

$At 300 K, one mole of a gas present in a 10 L flask exerted a pressure of 2.706 atm. What is its compressibility factor (Z)?$

1

$1.0$

2

$1.5$

3

$0.91$

4

$1.1$

Official Solution

Correct Option: $(3)$

$The compressibility factor is given by the formula: Where: - - - - - Substituting these values into the formula: Thus, the compressibility factor is 0.91.$

24

PYQ 2023

medium

chemistry ID: ap-eapce

$At T (K), equal weights of H, D, and T are present in closed vessel. The pressure exerted by this gaseous mixture is P atm. The ratio of partial pressures of T, D and H is approximately (H, D and T are isotopes of hydrogen).$

1

$0.33 : 0.33 : 0.33$

2

$0.18 : 0.27 : 0.54$

3

$0.25 : 0.50 : 0.25$

4

$0.54 : 0.27 : 0.18$

Official Solution

Correct Option: $(2)$

$At constant temperature, the pressure exerted by a gas in a closed vessel is proportional to the number of molecules present. Since equal weights of H, D, and T are given, the ratio of partial pressures depends on the molar masses. The ratio of partial pressures of gases with the same volume is inversely proportional to their molar masses. Thus, for equal masses: Given that the molar masses of H, D, and T are approximately in the ratio 1 : 2 : 3, the partial pressure ratio will be:$

25

PYQ 2023

medium

chemistry ID: ap-eapce

$The correct order of melting points of Al, Ga, In is:$

1

2

3

4

Official Solution

Correct Option: $(1)$

$The melting points of Al, Ga, and In are governed by their atomic structure and bonding characteristics. Based on the trends in metallic bonding: - (Gallium) has the lowest melting point among the three elements, primarily because of weaker metallic bonds and a relatively low atomic number. - (Indium) has a slightly higher melting point than Ga due to its larger atomic size and stronger metallic bonds. - (Aluminum) has the highest melting point because it exhibits stronger metallic bonding due to its smaller atomic size and higher charge density. Thus, the correct order is .$

26

PYQ 2023

medium

chemistry ID: ap-eapce

$If 100 L of an ideal gas at a pressure of 2 atm is compressed isothermally and reversibly to a final volume of ‘X’ L releases -460.6 L atm heat, the final volume ‘X’ (in L) is$

1

$1$

2

$10$

3

$100$

4

$2$

Official Solution

Correct Option: $(3)$

$For an isothermal and reversible process, the work done is given by: where is the number of moles, is the universal gas constant, is the temperature, is the final volume, and is the initial volume. We can use the first law of thermodynamics: where is the heat released, and is the change in internal energy. Since the process is isothermal,, so: Given: We use the formula for the work done: Substitute the given values: Solve for : Thus, the final volume is . Thus, the correct answer is .$

27

PYQ 2023

medium

chemistry ID: ap-eapce

$At 300 K the enthalpy change for the following reaction is -2800 kJ mol What is the for the same reaction at 300 K (kJ mol)?$

1

$+2802.49$

2

$-2800.00$

3

$-2814.94$

4

$+2802.49$

Official Solution

Correct Option: $(2)$

$For reactions involving gases, the relation between the enthalpy change () and internal energy change () is given by: Where: - is the enthalpy change - is the internal energy change - is the change in the number of moles of gases - is the gas constant (0.0821 L atm mol K) - is the temperature in Kelvin For the given reaction: - - Now we use the relation: Substitute the values: Thus, the value of is -2800.00 kJ/mol.$

28

PYQ 2023

medium

chemistry ID: ap-eapce

$The isotherms of an ideal gas at along with their slopes (m) (in the brackets) are shown here. If, then the correct order of slopes of these isotherms is:$

1

2

3

4

Official Solution

Correct Option: $(4)$

$Step 1: Understanding the relationship between isotherms and slopes. For an ideal gas, the equation of state is given by the ideal gas law: This implies that for a given temperature, the relationship between pressure and volume is inversely proportional. The slope of the isotherm, denoted as, is given by: At higher temperatures, the gas molecules have higher kinetic energy, leading to a steeper slope of the isotherm. Therefore, for, the isotherm at will have the steepest slope, followed by the isotherm at, and the isotherm at will have the smallest slope. Step 2: Analyzing the slopes. Since, the slope order will be: Thus, the correct order of the slopes of these isotherms is .$

29

PYQ 2023

medium

chemistry ID: ap-eapce

$If the standard Gibbs energy, enthalpy and entropy changes, and for a reaction are,, and respectively, the temperature of the reaction (in) is$

1

$350$

2

$427$

3

$77$

4

$35$

Official Solution

Correct Option: $(3)$

$We use the Gibbs free energy relation: Convert all to consistent units (J/mol): - J/mol - J/mol - J/K\cdotmol$

30

PYQ 2023

medium

chemistry ID: ap-eapce

$If the mean bond dissociation energies of C-C and C-H are 80 and 90 kJ mol respectively, the standard enthalpy of atomization of n-butane(g) (in kJ mol) is$

1

$-980$

2

$1040$

3

$1140$

4

$-1140$

Official Solution

Correct Option: $(3)$

$n-butane () structure: - 3 C-C bonds - 10 C-H bonds Given: - C-C bond energy = 80 kJ/mol - C-H bond energy = 90 kJ/mol$

31

PYQ 2023

medium

chemistry ID: ap-eapce

$Identify the correct statements from the following (A) The compressibility factor (Z) for an ideal gas is 1 (B) Uranium isotopes (and) are separated by converting them into vapours (C) Decrease in temperature increases the kinetic energy of gas molecules$

1

$A, B, C$

2

$A, C only$

3

$B, C only$

4

$A, B only$

Official Solution

Correct Option: $(4)$

$Let's analyze each statement: Option (A) **The compressibility factor (Z) for an ideal gas is 1. ** The compressibility factor is defined as, where is pressure, is volume, is the number of moles, is the ideal gas constant, and is temperature. For an ideal gas, the equation of state is, so . This statement is correct. Option (B) **Uranium isotopes (and) are separated by converting them into vapours. ** The separation of uranium isotopes and is industrially achieved using the gaseous diffusion method. Uranium is converted to uranium hexafluoride (), which is a volatile solid that sublimes to form a gas at relatively low temperatures. The slight mass difference between and molecules allows for their separation by passing the gaseous through a series of porous barriers. The lighter diffuses slightly faster than the heavier . This statement is correct. Option (C) **Decrease in temperature increases the kinetic energy of gas molecules. ** The average kinetic energy of gas molecules is directly proportional to the absolute temperature (in Kelvin). The relationship is given by, where is the Boltzmann constant. Therefore, a decrease in temperature leads to a decrease in the average kinetic energy of gas molecules, not an increase. This statement is incorrect. Based on the analysis, statements A and B are correct, while statement C is incorrect. Therefore, the correct option is (D) A, B only.$

32

PYQ 2023

medium

chemistry ID: ap-eapce

$If the standard enthalpy of sublimation of solid, naphthalene, Li and Na are 25.2, 73.0, 162, 108 kJ mol respectively, the order of sublimation of these substances with temperature is$

1

$Na>Li>Naphthalene>Solid$

2

$Solid >Na>Naphthalene>Li$

3

$Solid >Naphthalene>Na>Li$

4

$Li>Na>Naphthalene>Solid$

Official Solution

Correct Option: $(3)$

$The enthalpy of sublimation is the energy required to transform one mole of a substance from the solid phase directly to the gaseous phase at a constant temperature and pressure (usually standard conditions). A substance with a lower enthalpy of sublimation will require less energy to vaporize, meaning it will sublime more readily at a given temperature compared to a substance with a higher enthalpy of sublimation. Therefore, the order of sublimation with temperature will be inversely related to the enthalpy of sublimation. The substance with the lowest will sublime most readily at a given temperature, followed by the substance with the next lowest, and so on. Given the standard enthalpies of sublimation: - Solid : kJ mol - Naphthalene: kJ mol - Li: kJ mol - Na: kJ mol Arranging these values in increasing order: kJ mol This corresponds to the order: Solid e., the ease with which these substances sublime) will be the reverse of this order: Solid >Naphthalene>Na>Li Therefore, the correct order of sublimation of these substances with temperature is Solid >Naphthalene>Na>Li.$

33

PYQ 2023

medium

chemistry ID: ap-eapce

$Which one of the following options is correct for$

1

$Isothermal irreversible expansion of an ideal gas with constant external pressure.$

2

$Isothermal irreversible compression of an ideal gas with constant external pressure.$

3

$Isothermal reversible expansion of an ideal gas against constant external pressure.$

4

$Irreversible expansion of an ideal gas into vacuum.$

Official Solution

Correct Option: $(2)$

$When and, it implies that the process is an irreversible compression, meaning that the gas is compressed and work is done by the surroundings on the gas. The heat flow is negative (), which indicates that the gas is losing heat during the compression process. Therefore, this situation is described by an isothermal irreversible compression of an ideal gas at constant external pressure.$

34

PYQ 2023

medium

chemistry ID: ap-eapce

$11.0 L of an ideal gas at constant external pressure of 5 atm is compressed isothermally to a final volume of one liter. The heat absorbed and work done respectively, during this compression (in L atm) are$

1

$-50, -50$

2

$50, -50$

3

$-50, 50$

4

$50, 50$

Official Solution

Correct Option: $(3)$

$Work done, Since the process is isothermal for an ideal gas:$

35

PYQ 2023

medium

chemistry ID: ap-eapce

$Which of the following reaction is not feasible?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$The reaction involves the dehydration of ethanol to ethene, which is an example of an elimination reaction. The dehydration of alcohols occurs by the following mechanism: - For option (1), the reaction proceeds with concentrated at 443 K, which is a feasible condition for dehydration of ethanol to form ethene. - For option (3), dehydration also occurs in the presence of 85% at 440 K, which is again a feasible reaction. - For option (4), the reaction takes place with 20% at 358 K, which is also feasible under the given conditions. - For option (2), the reaction occurs with 75% at 300 K. However, at this relatively low temperature, the dehydration of ethanol to ethene is not favored, making this reaction non-feasible. Thus, the correct answer is option (2).$

36

PYQ 2023

medium

chemistry ID: ap-eapce

$The standard enthalpy of formation of (g), and are -393, -634, -1210 kJ mol respectively. If all the substans are in standard state, the standard enthalpy of decomposition of calcium carbonate to and (in kJ mol) is$

1

2

3

4

Official Solution

Correct Option: $(2)$

$We are given a decomposition reaction: To find the standard enthalpy change () for this decomposition, apply Hess's law: Given: Substitute into the formula: Hence, the standard enthalpy of decomposition is, indicating an endothermic reaction.$

37

PYQ 2023

medium

chemistry ID: ap-eapce

$On complete combustion of one mole of butane according to the equation: C H + O \to 4CO + 5H O, the standard enthalpy of the reaction is -2879 kJ/mol. If [CO] = -394, [H O] = -286, the standard enthalpy of formation of butane is$

1

$-3006$

2

$3006$

3

$127$

4

$-127$

Official Solution

Correct Option: $(4)$

$Use Hess's law: -2879 = [4(-394) + 5(-286)] - (butane) = [-1576 + (-1430)] = -3006 So, -2879 = -3006 - (butane) (butane) = -3006 + 2879 = -127 kJ/mol$

38

PYQ 2024

medium

chemistry ID: ap-eapce

$Identify the molecule for which the enthalpy of atomization and bond dissociation enthalpy are not equal.$

1

2

3

4

Official Solution

Correct Option: $(4)$

$To determine which molecule has different enthalpy of atomization and bond dissociation enthalpy, let's consider the definitions and examine each option: Enthalpy of Atomization: This is the enthalpy change when one mole of a compound in its standard state is converted into atoms. Bond Dissociation Enthalpy: This is the enthalpy change required to break a particular bond in a molecule in its gaseous state. For diatomic molecules like,, and, the process of converting to atoms and breaking a bond is essentially the same, as they consist of two atoms bound by a single covalent bond. However, in the case of (methane), the molecule consists of one carbon atom bonded to four hydrogen atoms: Here, the enthalpy of atomization involves breaking four C-H bonds to convert one mole of methane to its constituent atoms, which is more complex than a single bond dissociation. Thus, the enthalpy of atomization for will not be equal to the bond dissociation enthalpy of a single C-H bond, unlike diatomic molecules where both values would be equal. Therefore, the molecule for which and are not equal is .$

39

PYQ 2024

easy

chemistry ID: ap-eapce

$Observe the following reaction: . The standard enthalpy of formation of is:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$The given reaction is: . This indicates that the reaction releases 600 kJ of energy. We need to calculate the standard enthalpy of formation of . First, understand that the enthalpy change for the reaction,, is given as (negative because it is exothermic). The standard enthalpy change of the reaction is related to the standard enthalpy of formation of products and reactants by the formula: For the product, the enthalpy can be calculated as . For the reactants, since they are in their standard states, their standard enthalpies of formation are zero: Thus the equation simplifies to: Solving for : Therefore, the standard enthalpy of formation of is .$

40

PYQ 2024

medium

chemistry ID: ap-eapce

$Identify the correct equation relating,, and for 1 mole of an ideal gas (R = gas constant):$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Apply the First Law of Thermodynamics for an Ideal Gas For an ideal gas: Rearrange to solve for : Thus, the correct answer is .$

41

PYQ 2024

medium

chemistry ID: ap-eapce

$The number of extensive properties in the following list is: Enthalpy, density, volume, internal energy, temperature.$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Categorize the Properties - Extensive: Enthalpy, Volume, Internal Energy - Intensive: Density, Temperature Step 2: Count Extensive Properties$

42

PYQ 2025

medium

chemistry ID: ap-eapce

$For which of the following processes entropy change () is negative? I) Sublimation of dry ice II) Freezing of water III) Crystallisation of the dissolved substance IV) Burning of rocket fuel$

1

$I & II only$

2

$II & III only$

3

$III & IV only$

4

$I & IV only$

Official Solution

Correct Option: $(2)$

$Step 1: Understand entropy () concept. Entropy is a measure of disorder or randomness in a system. When a system becomes more ordered, entropy decreases (). Step 2: Analyze each process. I) Sublimation of dry ice: Solid gas — disorder increases II) Freezing of water: Liquid solid — particles become more ordered III) Crystallisation of dissolved substance: Ions/molecules settle into a lattice IV) Burning of rocket fuel: Produces gases and heat Conclusion: Only II and III lead to a decrease in entropy.$

43

PYQ 2025

medium

chemistry ID: ap-eapce

$The energy required to increase the temperature of 180 g of liquid water from 10 C to 15 C is 3765 J. What is of water in J mol K ? (= 18 u)$

1

$75.3$

2

$376.5$

3

$753$

4

$37.65$

Official Solution

Correct Option: $(1)$

$Given: Mass = 180 g,, Energy = 3765 J, Molar mass of water = 18 g/mol Number of moles = mol J mol K$

44

PYQ 2025

medium

chemistry ID: ap-eapce

$At (K), hydrogen and oxygen gases are mixed in the ratio of by mass in a closed vessel of volume litres. If the total pressure of the gaseous mixture is atm, the partial pressure of oxygen (in atm) is:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Determining the Number of Moles Given the mass ratio of hydrogen to oxygen is , let: - Mass of = g - Mass of = g The molar masses are: - g/mol - g/mol Thus, number of moles: Step 2: Computing Mole Fraction Total moles: Mole fraction of : Step 3: Finding Partial Pressure Using Dalton’s Law: Conclusion Thus, the correct answer is:$

45

PYQ 2025

medium

chemistry ID: ap-eapce

$Which of the following statements is incorrect about enthalpy?$

1

$Its absolute value can be determined accurately.$

2

$It is a state function.$

3

$It is an extensive property.$

4

$Enthalpy change can be determined using the first law of thermodynamics.$

Official Solution

Correct Option: $(1)$

$Step 1: Understand what enthalpy is. Enthalpy is a thermodynamic state function defined as: However, only changes in enthalpy ( ) can be measured, not the absolute value. Step 2: Evaluate each option. \begin{itemize} \item Option (1) is incorrect because the absolute value of enthalpy cannot be measured—only enthalpy changes are meaningful. \item Option (2) is correct: Enthalpy is a state function. \item Option (3) is correct: Enthalpy depends on the amount of substance, so it is extensive. \item Option (4) is correct: Enthalpy change relates to heat change under constant pressure, which is derived using the first law of thermodynamics. \end{itemize}$

46

PYQ 2025

medium

chemistry ID: ap-eapce

$Identify the correct statements I) Enthalpy is an intensive property II) For, H O(l) H O(g) process, increases III) Entropy is a state function$

1

$I, II, III$

2

$I, II only$

3

$I, III only$

4

$II, III only$

Official Solution

Correct Option: $(4)$

$Step 1: Evaluate Statement I - Enthalpy is an intensive property An intensive property is a physical property of a system that does not depend on the system's size or the amount of material in the system (e.g., temperature, density, pressure). An extensive property is a physical property that depends on the amount of matter in a sample (e.g., mass, volume, energy, enthalpy). Enthalpy (H) is a measure of the total heat content of a system. Its value depends on the amount of substance present. For example, 1 mole of a substance has a certain enthalpy, but 2 moles of the same substance will have twice that enthalpy. Therefore, enthalpy is an extensive property, not an intensive property. So, Statement I is Incorrect. Step 2: Evaluate Statement II - For, H O(l) H O(g) process, increases Entropy (S) is a measure of the disorder or randomness of a system. The process H O(l) H O(g) represents the phase transition from liquid water to gaseous water (vaporization). In the gaseous state, molecules are much more dispersed and have greater freedom of movement compared to the liquid state, where molecules are more closely packed and have less translational freedom. This increased disorder and randomness in the gaseous phase implies higher entropy. Therefore, for the transition from liquid to gas, the entropy of the system increases, meaning . So, Statement II is Correct. Step 3: Evaluate Statement III - Entropy is a state function A state function (or function of state) is a property whose value depends only on the current state of the system, not on the path taken to reach that state. Examples include temperature, pressure, volume, internal energy, enthalpy, and entropy. Entropy is a fundamental thermodynamic property. The change in entropy for a process depends only on the initial and final thermodynamic states of the system, irrespective of the specific path or process followed between these states. So, Statement III is Correct. Step 4: Identify the Correct Combination of Statements Based on our analysis: \begin{itemize} \item Statement I is Incorrect. \item Statement II is Correct. \item Statement III is Correct. \end{itemize} Therefore, the correct statements are II and III. Step 5: Analyze the Options \begin{itemize} \item Option (1): I, II, III. Incorrect because Statement I is false. \item Option (2): I, II only. Incorrect because Statement I is false. \item Option (3): I, III only. Incorrect because Statement I is false. \item Option (4): II, III only. Correct, as both Statement II and Statement III are true. \end{itemize}$

47

PYQ 2025

medium

chemistry ID: ap-eapce

$Identify the conditions at which van der Waals equation of state changes to ideal gas equation.$

1

$High temperature and high pressure$

2

$Low temperature and high pressure$

3

$High temperature and low pressure$

4

$Low temperature and low pressure$

Official Solution

Correct Option: $(3)$

$Step 1: Understand van der Waals equation The van der Waals equation accounts for molecular volume and intermolecular forces, making it more accurate than ideal gas law. Step 2: Conditions for ideal gas approximation At high temperature, molecules move faster, reducing effects of intermolecular forces. At low pressure, molecules are far apart, making volume of molecules negligible. Step 3: Conclusion Hence, van der Waals equation approaches ideal gas law at high temperature and low pressure.$

48

PYQ 2025

medium

chemistry ID: ap-eapce

$The enthalpy change (ΔH) for the reaction N 2 (g) + 3H 2 (g) \to 2NH 3 (g) is -92 kJ/mol. What is the enthalpy change for the decomposition of 1 mol of NH 3 (g) into its elements?$

1

2

3

4

$\( -92 \, \text{kJ/mol}$

Official Solution

Correct Option: $(1)$

$Given Reaction: N 2 (g) + 3H 2 (g) \to 2NH 3 (g), ΔH = -92 kJ/mol. This ΔH corresponds to the formation of 2 moles of NH 3 . Enthalpy for Formation of 1 Mole of NH 3 : For 1 mole of NH 3, the reaction is ½N 2 (g) + 1½H 2 (g) \to NH 3 (g), so ΔH = -92 / 2 = -46 kJ/mol. Reverse Reaction (Decomposition): The decomposition reaction is NH 3 (g) \to ½N 2 (g) + 1½H 2 (g). Since enthalpy is a state function, the enthalpy change for the reverse reaction is the negative of the forward reaction: ΔH decomp = -(-46) = +46 kJ/mol. Conclusion: The enthalpy change for the decomposition of 1 mole of NH 3 is +46 kJ/mol .$

49

PYQ 2025

medium

chemistry ID: ap-eapce

$The standard enthalpy of formation of C H (g), CO (g), and H O(l) are 52, -394, and -286 kJ mol respectively. The heat evolved by burning 7 g of C H (g) is$

1

$353 kJ$

2

$1412 kJ$

3

$706 kJ$

4

$1306 kJ$

Official Solution

Correct Option: $(3)$

$Step 1: Write the balanced combustion reaction. Step 2: Use enthalpies of formation to calculate Step 3: Find moles of C H in 7 g Step 4: Multiply with enthalpy change per mole$

50

PYQ 2025

medium

chemistry ID: ap-eapce

$At K, and for the following reaction are kJ and J; . What is (in kJ) for this reaction? (J mol)$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Gibbs Free Energy Formula The Gibbs free energy change is given by: Since and are related by: Here, . Thus, Step 2: Computing Conclusion Thus, the correct answer is:$

51

PYQ 2025

medium

chemistry ID: ap-eapce

$A and B are ideal gases. At T(K), 2 L of A with a pressure of 1 bar is mixed with 4 L of B with a pressure of P bar in a 100 L flask. The pressure exerted by gaseous mixture is 0.1 bar. What is the value of P in bar?$

1

$0.04$

2

$0.02$

3

$0.10$

4

$0.05$

Official Solution

Correct Option: $(2)$

$We can apply Dalton's Law of partial pressures. The total pressure exerted by the mixture is the sum of the partial pressures of each gas: Using the ideal gas law,, we can calculate the partial pressures. For gas A: For gas B, the partial pressure is calculated similarly, and we find that the value of bar.$

52

PYQ 2025

medium

chemistry ID: ap-eapce

$One mole of ethanol (C₂H₅OH) was completely burnt in oxygen to form CO₂(g) and H₂O(l). What is the (in kJ/mol) for this reaction?$

1

$1921$

2

$-1921$

3

$1367$

4

$-1367$

Official Solution

Correct Option: $(4)$

$The standard enthalpy change of formation for ethanol, CO₂, and H₂O is given. Using Hess's Law, the enthalpy of the combustion reaction is calculated by subtracting the enthalpy of formation of the reactants from the products. The correct value of is -1367 kJ/mol.$

53

PYQ 2025

medium

chemistry ID: ap-eapce

$At (K), hydrogen and oxygen gases are mixed in the ratio of by mass in a closed vessel of volume litres. If the total pressure of the gaseous mixture is atm, the partial pressure of oxygen (in atm) is:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Determining the Number of Moles Given the mass ratio of hydrogen to oxygen is , let: - Mass of = g - Mass of = g The molar masses are: - g/mol - g/mol Thus, number of moles: Step 2: Computing Mole Fraction Total moles: Mole fraction of : Step 3: Finding Partial Pressure Using Dalton’s Law: Conclusion Thus, the correct answer is:$

54

PYQ 2025

medium

chemistry ID: ap-eapce

$For one mole of an ideal gas, three isochors were obtained at and respectively. Their slopes are and . If, then the correct relationship of slopes is:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$For an ideal gas, at constant volume (isochor), pressure varies linearly with temperature : Slope of isochor in - graph is: Given and constants, slope inversely proportional to volume . Since,$

55

PYQ 2025

medium

chemistry ID: ap-eapce

$For one mole of an ideal gas an isochore is obtained. The slope of the isochore is . What will be its pressure (in atm) when the temperature is 12.2 K? (R =).$

1

$10.0$

2

$0.1$

3

$1.0$

4

$0.5$

Official Solution

Correct Option: $(3)$

$An isochore represents a process at constant volume (isochoric process). For an ideal gas, . Since the volume V is constant and mole, we can write . This shows that P is directly proportional to T for an isochoric process. The graph of P vs T is a straight line passing through the origin (if T is in Kelvin). The slope of this line (isochore on a P-T diagram) is . Given slope . So, . We are given mole and . This implies . Now, we need to find the pressure P when the temperature K. Using : . . . . So, atm. Alternatively, using with L, mol,, K: This matches option (3).$

56

PYQ 2025

medium

chemistry ID: ap-eapce

$Consider the following: Statement-I: Both internal energy (U) and work (w) are state functions. Statement-II: During the free expansion of an ideal gas into vacuum, the work done is zero. The correct answer is$

1

$Both Statement-I and Statement-II are correct$

2

$Both Statement-I and Statement-II are not correct$

3

$Statement-I is correct, but Statement-II is not correct$

4

$Statement-I is not correct, but Statement-II is correct$

Official Solution

Correct Option: $(4)$

$Let’s break this down step by step to evaluate the statements and determine why option (4) is the correct answer. Step 1: Evaluate Statement-I Internal energy () is a state function because it depends only on the state of the system. Work () is a path function because it depends on the process. Statement-I is incorrect because work is not a state function. Step 2: Evaluate Statement-II During free expansion of an ideal gas into a vacuum: External pressure . Work done, . Statement-II is correct. Step 3: Confirm the correct answer Statement-I is incorrect, but Statement-II is correct, matching option (4). Thus, the correct answer is (4) Statement-I is not correct, but Statement-II is correct.$

57

PYQ 2025

medium

chemistry ID: ap-eapce

$The enthalpy of formation of and are and respectively. If the heat of combustion of is, the enthalpy of formation of is$

1

$+216.5$

2

$-216.5$

3

$-355.5$

4

$+355.5$

Official Solution

Correct Option: $(2)$

$Step 1: Write the Balanced Chemical Equation for the Combustion of The combustion of methanol ( ) involves its reaction with oxygen ( ) to produce carbon dioxide ( ) and water ( ). Step 2: State the Formula for Enthalpy of Reaction from Enthalpies of Formation The enthalpy of a reaction ( ) can be calculated from the standard enthalpies of formation ( ) of the products and reactants using the following formula: Step 3: List the Given Enthalpies of Formation and Reaction Given: \begin{itemize} \item Enthalpy of formation of , \item Enthalpy of formation of , \item Heat of combustion of , \item Enthalpy of formation of is 0 (as it's an element in its standard state). \end{itemize} Let the enthalpy of formation of be . Step 4: Substitute the Values into the Enthalpy of Reaction Formula Using the balanced equation from Step 1 and the formula from Step 2: Substitute the known values: Step 5: Solve for Rearrange the equation to solve for : Step 6: Analyze Options \begin{itemize} \item Option (1): +216.5. Incorrect. \item Option (2): -216.5. Correct, as it matches our calculated enthalpy of formation. \item Option (3): -355.5. Incorrect. Option (4): +355.5. Incorrect. \end{itemize}$

58

PYQ 2025

medium

chemistry ID: ap-eapce

$Identify the correct statements about the following stoichiometric equation. [I.] [II.] [III.] The oxidation state of P in is$

1

$I, II, III$

2

$I, II only$

3

$I, III only$

4

$II, III only$

Official Solution

Correct Option: $(4)$

$Step 1: Analyze the given chemical equation The given reaction is: Step 2: Understand the balancing conditions From stoichiometric balancing: - Statement I: is incorrect upon checking the balanced equation. - Statement II: is found to be valid based on balancing. - Statement III: In, using oxidation number method: So, the oxidation state of phosphorus is, making the statement true. Step 3: Conclusion Only statements II and III are correct.$

59

PYQ 2025

medium

chemistry ID: ap-eapce

$The signs of and for a reaction to be spontaneous at all temperatures respectively are$

1

$positive, positive$

2

$positive, negative$

3

$negative, negative$

4

$negative, positive$

Official Solution

Correct Option: $(4)$

$Let’s break this down step by step to determine the signs of and for a reaction to be spontaneous at all temperatures and why option (4) is the correct answer. Step 1: Understand the condition for spontaneity The Gibbs free energy change determines spontaneity: A reaction is spontaneous if at all temperatures. Step 2: Analyze the signs of and If is negative (exothermic) and is positive (entropy increases): is always negative. Other combinations don’t ensure spontaneity at all temperatures. Step 3: Confirm the correct answer For spontaneity at all temperatures, must be negative and must be positive, matching option (4). Thus, the correct answer is (4) negative, positive.$

60

PYQ 2025

medium

chemistry ID: ap-eapce

$At 256 K, the rms speed of molecules is 3.16 m/s. What is the most probable velocity (in m/s) of the same gas molecules at the same temperature?$

1

$2.911$

2

$2.58$

3

$3.16$

4

$1.29$

Official Solution

Correct Option: $(2)$

$For an ideal gas, the root mean square speed is, and the most probable speed is . Given m/s: This matches option (2).$

61

PYQ 2025

medium

chemistry ID: ap-eapce

$If and are standard enthalpy change and standard entropy change respectively for a reaction, the incorrect option is$

1

$= negative; = positive; spontaneous at all temperatures$

2

$= negative; = negative; non-spontaneous at low temperatures$

3

$= positive; = positive; non-spontaneous at low temperatures$

4

$= negative; = negative; spontaneous at low temperatures$

Official Solution

Correct Option: $(2)$

$Gibbs free energy determines spontaneity: . (1), : will always be negative, so spontaneous at all temperatures. (2), : At low temperatures, the term will be small, so can be negative (spontaneous). As temperature increases, the term becomes larger, and may become positive (non-spontaneous). The option says "non-spontaneous" at low T, so this is incorrect. (3), : At low temperatures, the term is small, and will be positive (non-spontaneous). At high temperatures, can become larger than, making negative (spontaneous). (4), : At low temperatures, the reaction can be spontaneous. This statement in itself isn't necessarily incorrect, but it can be depending on the magnitude of and compared to the temperature. Since we are looking for the "incorrect" option, (2) is definitely incorrect. (4) can be incorrect as well depending on the context. (4) can be considered a better answer if it means "always spontaneous" at low T, which is false.$

62

PYQ 2025

medium

chemistry ID: ap-eapce

$The C of H O(l) is 75.3 J mol K . What is the energy (in J) required to raise 180 g of liquid water from 10 C to 15 C? (H O = 18 u)$

1

$3.765$

2

$3765$

3

$753$

4

$376.5$

Official Solution

Correct Option: $(2)$

$The heat () required to raise the temperature of a substance is given by, where is the mass, is the specific heat capacity, and is the change in temperature. However, we are given the molar heat capacity (C), so we need to use the number of moles (). The relationship is . First, calculate the number of moles of water: $ $$

63

PYQ 2025

medium

chemistry ID: ap-eapce

$For one mole of an ideal gas an isochore is obtained. The slope of the isochore is . What will be its pressure (in atm) when the temperature is 12.2 K? (R =).$

1

$10.0$

2

$0.1$

3

$1.0$

4

$0.5$

Official Solution

Correct Option: $(3)$

$An isochore represents a process at constant volume (isochoric process). For an ideal gas, . Since the volume V is constant and mole, we can write . This shows that P is directly proportional to T for an isochoric process. The graph of P vs T is a straight line passing through the origin (if T is in Kelvin). The slope of this line (isochore on a P-T diagram) is . Given slope . So, . We are given mole and . This implies . Now, we need to find the pressure P when the temperature K. Using : . . . . So, atm. Alternatively, using with L, mol,, K: This matches option (3).$

64

PYQ 2025

medium

chemistry ID: ap-eapce

$At K, and for the following reaction are kJ and J; . What is (in kJ) for this reaction? (J mol)$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Gibbs Free Energy Formula The Gibbs free energy change is given by: Since and are related by: Here, . Thus, Step 2: Computing Conclusion Thus, the correct answer is:$

65

PYQ 2025

medium

chemistry ID: ap-eapce

$One mole of was completely burnt in oxygen to form and . The standard enthalpy of formation () of, and is x, y, z kJ mol respectively. What is (in kJ mol) for this reaction?$

1

2

3

4

Official Solution

Correct Option: $(4)$

$The combustion reaction for ethanol () is: Balance the equation: Carbons: 2 on left, so 2 CO on right. Hydrogens: 5+1=6 on left, so 3 H O on right. Oxygens: On right, oxygen atoms. On left, 1 oxygen in C H OH. Need 6 more from O . So, 3 O . Balanced equation: The standard enthalpy of reaction () is calculated using standard enthalpies of formation () of products and reactants: Given standard enthalpies of formation: kJ/mol kJ/mol kJ/mol The standard enthalpy of formation of an element in its standard state is zero. So, . Now, calculate : This can be written as . This matches option (4).$

66

PYQ 2025

medium

chemistry ID: ap-eapce

$At 298 K, for the reaction is 165.469 kJ mol . What is the equilibrium constant for this reaction?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Use Gibbs free energy relation where, . Step 2: Calculate Step 3: Calculate equilibrium constant Step 4: Conclusion Equilibrium constant is .$

67

PYQ 2025

medium

chemistry ID: ap-eapce

$Consider the following equilibrium reaction in gaseous state at T(K): The initial concentration of B is 1.5 times that of A. At equilibrium, the concentrations of A and B are equal. The equilibrium constant for the reaction is:$

1

$6$

2

$16$

3

$12$

4

$4$

Official Solution

Correct Option: $(4)$

$Step 1: Let initial concentration of A be Then, initial concentration of B is Step 2: Let change in concentration of A be Then, Step 3: Use equilibrium condition Given: \text{[A] = [B]} at equilibrium So, Step 4: Calculate equilibrium concentrations Step 5: Apply equilibrium expression Step 6: Conclusion The equilibrium constant$

68

PYQ 2025

medium

chemistry ID: ap-eapce

$The enthalpies of formation of,, and are,, and, respectively. The quantity of heat liberated when of is burnt completely in oxygen is.}$

1

2

3

4

Official Solution

Correct Option: $(3)$

$Step 1: Known Information. Enthalpy of formation of : Enthalpy of formation of : Enthalpy of formation of : Mass of : Step 2: Balanced Combustion Equation. The balanced combustion equation for glucose ( ) is:$ ) is given by:\) : : (standard state reference)Products: : : Substitute into the equation: \) ) is: \) Number of moles of glucose:\) :\) $ Final Answer:$

69

PYQ 2025

medium

chemistry ID: ap-eapce

$Identify the incorrect statements from the following. I. For an adiabatic process, . II. Enthalpy is an intensive property. III. For the process, the entropy increases.$

1

$I only$

2

$I, II, III$

3

$I, III only$

4

$II, III only$

Official Solution

Correct Option: $(4)$

$Step 1: Analyze Statement I. Statement I: For an adiabatic process, . In an adiabatic process, there is no heat exchange with the surroundings ( ). The first law of thermodynamics states: $ ): \) ) is defined as: \) $ where: is internal energy, is pressure, is volume. Both internal energy ( ) and the product are extensive properties (they depend on the amount of substance). Therefore, enthalpy is an extensive property, not an intensive property. This statement is incorrect. Step 3: Analyze Statement III. Statement III: For the process , the entropy increases. Entropy is a measure of disorder or randomness. When water transitions from liquid ( ) to solid ( ), the molecules become more ordered (crystalline structure in ice). This decrease in disorder means that the entropy decreases during this process. This statement is incorrect. Step 4: Identify Incorrect Statements. From the analysis: Statement I is correct. Statement II is incorrect. Statement III is incorrect. Thus, the incorrect statements are II and III. Final Answer:$

70

PYQ 2025

medium

chemistry ID: ap-eapce

$Identify the correct statements from the following: I) Work is a path function. II) Enthalpy is an extensive property. III) Lattice enthalpy of ionic compounds can be obtained from Born-Haber cycle.$

1

$I only$

2

$I, III only$

3

$II, III only$

4

$I, II, III$

Official Solution

Correct Option: $(4)$

$Step 1: Analyze each statement. Statement I: True. Work depends on the path taken and not just the initial and final states, so it is a path function. Statement II: True. Enthalpy is proportional to the amount of substance, hence it is an extensive property. Statement III: True. Lattice enthalpy can be determined using the Born-Haber cycle by applying Hess's law. Conclusion: All three statements are correct.$

High-Yield Trend

Chapter Questions 70 MCQs

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About Thermodynamics - AP-EAPCET

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Chapter Questions
70 MCQs