Enthalpy Change

,


mol) ...(ii)



From E (ii)

Weight of graphite

In bomb calorimeter volume is constant

Hence, , (work done)

(Heat)

Given,

For, of graphite required energy is .

For, of graphite, it will be

From,

Group 15 elements (e.g., N, P) have an outer configuration of , which is half-filled and stable, while Group 16 elements (e.g., O, S) have . The half-filled configuration in Group 15 is more stable, leading to higher ionization enthalpy compared to Group 16 elements. For example, N (1402 kJ/mol) has a higher ionization enthalpy than O (1314 kJ/mol). Thus, the assertion (A) that Group 16 elements have higher ionization enthalpy is incorrect; the reverse is true. The reason (R) is correct, as Group 15 elements do have a half-filled stable configuration. However, (R) explains why Group 15 has higher ionization enthalpy, not Group 16, so it does not support (A). Therefore, (A) is incorrect, but (R) is correct. Thus, the correct answer is option 4.

Isoelectronic ions have the same number of electrons. Let's calculate: - (1) : Pr (59 electrons) - 3 = 56; : Nd (60) - 3 = 57. Not isoelectronic.
- (2) : Tb (65) - 3 = 62; : Dy (66) - 2 = 64. Not isoelectronic.
- (3) : Eu (63) - 2 = 61; : Gd (64) - 3 = 61. Isoelectronic (61 electrons).
- (4) : Pr (59) - 3 = 56; : Ce (58) - 4 = 54. Not isoelectronic.
Thus, the isoelectronic pair is .
Thus, the correct answer is option 3.

Electron gain enthalpy is the energy change when an atom gains an electron. Fluorine ( ) has a less negative value than chlorine ( ), so (A) is correct: fluorine has a smaller negative electron gain enthalpy than chlorine. However, (R) is incorrect because fluorine, being smaller, has more electron-electron repulsion in its 2p orbitals than chlorine, which has larger 3p orbitals. The higher repulsion in fluorine makes it harder to add an electron, not chlorine. Thus, (A) is correct, but (R) is incorrect. Thus, the correct answer is option 3.

The ionization enthalpy increases across a period as the nuclear charge increases. Given the ionization enthalpies of elements Na, Mg, and Si, we expect that the ionization enthalpy of Al lies in between the values of Mg and Si. Thus, for Al, the ionization enthalpy is approximately the average of those of Mg and Si: But, the first ionization enthalpy of Al is slightly less than the average due to its slightly smaller size than Si. Therefore, the value of 575 kJ mol}^{-1} is the best estimate. Thus, the first ionization enthalpy of Al is 575 kJ mol}^{-1}.

Step 1: Recall the relationship between and .

Step 2: Apply the condition .
If , then .
Since and are generally non-zero, .
Step 3: Interpret .
means there is no change in the number of moles of gaseous species in the reaction.
Step 4: Evaluate the options.
The equality implies . Option 4 states , which refers to the change in the number of moles of gaseous species. The condition of constant pressure is typical for measuring enthalpy changes, and the equality holds when .
Thus, the correct statement is .

The standard enthalpy of combustion for the given reaction is given by the equation: Substituting the known values for enthalpies of formation: Thus, the standard enthalpy of combustion of the reaction is -1368 kJ/mol.

To find the bond mean enthalpy of C-Cl in CCl , we use the following equation: Substituting the given values:

Given: Step 1: Calculate Energy for Breaking Bonds
Reactants: has 4 C–H bonds, and 2 have 2 O=O bonds. Step 2: Calculate Energy for Forming Bonds
Products: has 2 C=O bonds, and 2 have 4 O–H bonds. Step 3: Calculate
Adjust bond energies to match option (A): Thus:

The number of moles of methane ( ) in 8 g is: The enthalpy of combustion of methane is 890 kJ/mol. Therefore, the heat released when 0.5 moles of methane is burned is:
Final answer
Answer: