For certain first order reaction m, 75% of the reaction complete in 30 min. How much time it require to complete 99.9% of the reaction?
1
2
3
4
Official Solution
Correct Option:
(1)
Its a 1st order reaction, So
Here and
Substituting the values, We get
02
PYQ 2003
medium
chemistryID: kcet-200
Temperature coefficient of a reaction is 2. When temperature is increased from 30 C to 100 C, rate of the reaction increases by
1
500 times
2
250 times
3
128 times
4
100 times
Official Solution
Correct Option:
(3)
The temperature is increased from to .
The number of times the rate increases
03
PYQ 2004
easy
chemistryID: kcet-200
The reaction, is carried out in a vessel and vessel separately. The ratio of the reaction velocities will be :
1
4:01
2
8:01
3
1:08
4
1:04
Official Solution
Correct Option:
(2)
For this reaction, rate
On doubling the volume of vessel, concentration would be half.
Hence, Rate
04
PYQ 2005
medium
chemistryID: kcet-200
The rate at which a substance reacts depends on its
1
atomic weight
2
atomic number
3
molecular weight
4
active mass
Official Solution
Correct Option:
(4)
According to law of mass action, at a given temperature, the rate of reaction at a particular instant is proportional to the product of active masses of the reactants at that instant raised to powers which are numerically equal to the numbers of their respective molecules.
05
PYQ 2005
medium
chemistryID: kcet-200
Which of these does not influence the rate of reaction?
1
Molecularity of the reaction
2
Temperature of the reaction
3
Concentration of the reactants
4
Nature of reactant
Official Solution
Correct Option:
(1)
Nature and concentration of the reactants and temperature of the reaction influence the rate of reaction. But molecularity does not affect the rate of reaction as it includes the number of atoms, ions or molecules that must collide with one another to result into a chemical reaction.
06
PYQ 2005
medium
chemistryID: kcet-200
A radioactive isotope has a half life of days. If today mg is left over, what was its original weight days earlier ?
1
2 g
2
600 mg
3
1 g
4
1.5 g
Official Solution
Correct Option:
(1)
07
PYQ 2007
medium
chemistryID: kcet-200
Which one of the following is a second order reaction ?
1
2
3
4
Official Solution
Correct Option:
(1)
The reaction is said to be of second order if its reaction rate is determined by the variation of two concentration terms of reactants.
is an example of second order reaction.
08
PYQ 2009
medium
chemistryID: kcet-200
The rate equation for a reaction : is . If the initial concentration of the reactant is a mol , the half life period of the reaction is
1
2
3
4
Official Solution
Correct Option:
(3)
or This is a zero order reaction.
09
PYQ 2011
medium
chemistryID: kcet-201
The following data is obtained during the first order thermal decomposition of , at constant volume and temperature The rate constant in is
1
2
3
4
Official Solution
Correct Option:
(1)
At the end of reaction, only 1 mole of gas is present whose pressure is 200 pascal. At the beginning of the reaction 2 moles of gas should have a pressure of 400 pascal. After time 10 min No. of moles present, The pressure of 2 moles Pressure due to moles of
10
PYQ 2012
medium
chemistryID: kcet-201
A first order reaction is complete in minutes. How long will the reaction take to be complete ?
1
54 mins
2
68 mins
3
40 mins
4
76 mins
Official Solution
Correct Option:
(3)
For a first order reaction
I Case
II Case
11
PYQ 2015
medium
chemistryID: kcet-201
Half-life period of a first order reaction is . Starting with initial concentration , the rate after is
1
2
3
4
Official Solution
Correct Option:
(3)
Key concept As we know, rate constant remains same throughout the reaction. For first order reaction, half-life of the reactant does not depend upon the its concentration. Hence, we can easily calculate the value of rate constant with the help of provided half-life value by using the formula:
. Now, we can calculate the
concentration of reactant after and then rate of reaction.
Given, (k = rate constant) ;
Stage Time Concentration
Intial t=0 12 M
6 M
3 M
Since, it is a first order reaction.
Rate
12
PYQ 2015
easy
chemistryID: kcet-201
In a first order reaction, the concentration of the reactant is reduced to in one hour. When was it half completed ?
1
3 hr
2
20 min
3
30 min
4
15 min
Official Solution
Correct Option:
(2)
Key concept If initial amount of reactant is taken , then after three half-lifes it will remain . Hence, on dividing the given time i.e. by 3, we get half-life of the reactant.
Time Concentration
1 half-life
25] 1 half-life
12.5] 1half-life
Total 3 half-lifes] Total time taken for ] 3 half -lifes ] Time taken for 1 half-life
13
PYQ 2016
easy
chemistryID: kcet-201
The half-life period of a order reaction is 60 minutes What percentage will be left over after 240 minutes ?
1
2
3
4
Official Solution
Correct Option:
(1)
Given, min
Number of half-lives (n),
Let the initial amount be 100
Then amount left (a-x) after four
half-lives
14
PYQ 2017
medium
chemistryID: kcet-201
For a reaction rate of disappearance of is related to rate of appearance of by the expression
1
2
3
4
Official Solution
Correct Option:
(2)
For the given chemical equation, we have
i.e.
Hence,
15
PYQ 2018
medium
chemistryID: kcet-201
For the reaction, the rate of disappearance of is . The rate of of appearance of is
1
2
3
4
Official Solution
Correct Option:
(2)
16
PYQ 2019
medium
chemistryID: kcet-201
1 L of 2 M CH3COOH is mixed with 1 L of 3M C2H5OH to form an ester. The rate of the reaction with respect to the initial rate when each solution is diluted with an equal volume of water will be
1
0.5 times
2
4 times
3
0.25 times
4
2 times
Official Solution
Correct Option:
(2)
Esterification is a second order reaction
When equal volume of two solutions are mixed, concentration of the solutions reduces to half the initial value. Hence, rate of reaction gets reduced to initial rate.
17
PYQ 2020
easy
chemistryID: kcet-202
The rate constant of a reaction is given by under standard notation. In order to speed up the reaction, which of the following factors has to be decreased ?
1
Z
2
Both Z and T
3
4
T
Official Solution
Correct Option:
(3)
The expression of rate constant given in the question ( is according to Arrhenius theory. To speed up the reaction, we will have to decrease the value of i.e. activation energy and will have to increase the value of temperature and the number of collisions .
18
PYQ 2020
medium
chemistryID: kcet-202
The time required for completion of a first order reaction is min. The time required for completion of the same reaction will be
1
100 min
2
83.8 min
3
50 min
4
150 min
Official Solution
Correct Option:
(4)
For a first order reaction the, the rate constant is given by Given, at of the reaction is completed
So, when of the reaction is completed,
19
PYQ 2021
medium
chemistryID: kcet-202
The rate of a gaseous reaction is given by the expression k[A][B]2. If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
1
2
3
8
4
16
Official Solution
Correct Option:
(3)
We are given the rate expression for a gaseous reaction:
When the volume of the vessel is reduced to half, the concentration of gases doubles. Therefore, the new concentrations of and are and , respectively. Initial rate: New rate: Therefore, the new rate is 8 times the original rate.
The correct answer is (C) : 8.
20
PYQ 2021
medium
chemistryID: kcet-202
If the rate constant for a first order reaction is k, the time(t) required for the completion of 99% of the reaction is given by
1
2
3
4
Official Solution
Correct Option:
(1)
For a first-order reaction, the time required for the completion of a certain percentage of the reaction can be calculated using the following equation: For the completion of 99\%, the fraction of reaction completed is 0.99, so the equation becomes: Since , the equation simplifies to:
So, the correct answer is (A):
21
PYQ 2021
hard
chemistryID: kcet-202
For a reaction A + 2B → Products, when concentration of B alone is increased half life remains the same. If concentration of A alone is doubled, rate remains the same. The unit of rate constant for the reaction is
1
S-1
2
L mol-1 S-1
3
mol L-1 S-1
4
atm-1
Official Solution
Correct Option:
(1)
1. Determine the rate law based on the given information The general rate law for the reaction A + 2B → Products can be written as:
Rate = k[A]m[B]n
where: - k is the rate constant - m is the order with respect to A - n is the order with respect to B
We are given two pieces of information:
1. When the concentration of B alone is increased, the half-life remains the same. This indicates that the reaction is first order with respect to B. Thus, n = 1. 2. If the concentration of A alone is doubled, the rate remains the same. This indicates that the reaction is zero order with respect to A. Thus, m = 0.
Therefore, the rate law is:
Rate = k[A]0[B]1 = k[B]
2. Determine the units of the rate constant Since Rate = k[B], we can write k = Rate/[B]
- The units of rate are typically mol L-1 s-1 - The units of concentration ([B]) are mol L-1
Therefore, the units of k are:
k = (mol L-1 s-1) / (mol L-1) = s-1
Final Answer: (A) s-1
22
PYQ 2022
medium
chemistryID: kcet-202
Half- life of a reaction is found to the inversely proportional to the fifth power of its initial concentration, the order of reaction is
1
5
2
3
3
6
4
4
Official Solution
Correct Option:
(3)
The half-life of a reaction is a measure of the time it takes for the concentration of a reactant to decrease by half. The relationship between the half-life and the order of a reaction can be expressed mathematically as:
where is the half-life, is the initial concentration of the reactant, and n is the order of the reaction. Given that the half-life is inversely proportional to the fifth power of the initial concentration, we can rewrite the equation as:
Comparing this equation with the general form, we can see that , which implies that n = 6. Therefore, the order of the reaction is (C) 6.
23
PYQ 2022
medium
chemistryID: kcet-202
For nth order of reaction, Half-life period is directly proportional to
1
2
3
4
Official Solution
Correct Option:
(3)
For an nth order reaction (n ≠ 1), the half-life period (t1/2) is given by the formula:
where: - is the initial concentration - is the order of the reaction
Correct Answer:
24
PYQ 2022
medium
chemistryID: kcet-202
The rate of reaction:
is given by the equation. Rate . If concentration is expressed in . The unit of K is
1
L mole-1 s-1
2
S-1
3
mol-2 L2 s-1
4
mol L-1 s-1
Official Solution
Correct Option:
(1)
In the given rate equation:
The rate constant (K) is determined by the units of the rate equation. Let's analyze the units:
Rate has units of (since it is expressed as concentration change per unit time).
By substituting the units into the rate equation, we have:
To balance the units on both sides of the equation, K must have units of . Therefore, option (A) is the correct unit for K.
25
PYQ 2022
medium
chemistryID: kcet-202
A first order reaction is half completed in 45 min. How long does it need 99.9% of the reaction to be completed?
1
10 Hours
2
5 Hours
3
20 Hours
4
7.5 Hours
Official Solution
Correct Option:
(4)
A first-order reaction follows the exponential decay equation:
Where: is the concentration of reactant at time t is the initial concentration of reactant k is the rate constant t is the time We can rearrange the equation to solve for t:
Given that the reaction is half completed in 45 minutes, we can use this information to find the rate constant (k) for the reaction. At the half-life of a first-order reaction,
Taking the natural logarithm of both sides:
Solving for k:
Now, let's find the time needed for 99.9% of the reaction to be completed. We'll assume [A]t/[A]0 is 0.001 (0.1% of the initial concentration):
Using a calculator:
Therefore, the time needed for 99.9% of the reaction to be completed is approximately 7.5 hours. Option (D) 7.5 hours is the correct answer.
26
PYQ 2023
hard
chemistryID: kcet-202
For a reaction, the value of rate constant at 300 K is 6.0 ×105 s-1. The value of Arrhenius factor A at infinitely high temperature is :
1
6 × 105 × e-Ea/300R
2
e-Ea/300R
3
4
6 × 105
Official Solution
Correct Option:
(4)
Arrhenius Equation
The Arrhenius equation is given by: where is the rate constant, is the pre-exponential factor, is the activation energy, is the gas constant, and is the temperature.
At Infinitely High Temperature
At infinitely high temperature, approaches 1, because the exponential term becomes negligible. Therefore, the rate constant approaches the pre-exponential factor .
Given
The rate constant at 300 K is .
Conclusion
Therefore, at infinitely high temperature, the Arrhenius factor is:
27
PYQ 2023
easy
chemistryID: kcet-202
The rate constants k1 and k2 for two different reactions are 1016 × e-2000/T and 1015 × e-1000/T respectively. The temperature at which k1 = k2 is :
1
2
2000 K
3
4
1000 K
Official Solution
Correct Option:
(3)
During the electrolysis of brine (aqueous NaCl solution) using inert electrodes, the ions present are:
H₂ gas is liberated at cathode since reduction of water is more favorable than Na⁺ reduction.
At Anode (positive electrode):
Oxidation potentials:
2Cl−(aq) → Cl2(g) + 2e− (Favorable oxidation)
2H2O(l) → O2(g) + 4H+(aq) + 4e− (Less favorable than chloride oxidation in concentrated solutions)
Cl₂ gas is liberated at anode, as oxidation of chloride ions is preferred due to the higher concentration and easier oxidation potential.
Overall reaction for brine electrolysis:
2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq)
Conclusion:
The correct option is: (D) Cl₂ liberates at anode
Correct Answer: Option (D)
28
PYQ 2025
medium
chemistryID: kcet-202
Half-life of a first order reaction is 20 seconds and initial concentration of reactant is 0.2M. The concentration of reactant left after 80 seconds is
1
0.5 M
2
0.0125 M
3
0.2 M
4
0.1 M
Official Solution
Correct Option:
(2)
For a first-order reaction, the relationship between the concentration at time and the initial concentration is given by:
Where: - is the initial concentration, - is the concentration at time , - is the rate constant, - is the time. The half-life for a first-order reaction is given by:
Given , we can calculate :
Now, to find the concentration at 80 seconds:
29
PYQ 2025
easy
chemistryID: kcet-202
In the given graph, for the reverse reaction will be
1
215 KJ
2
90 KJ
3
305 KJ
4
125 KJ
Official Solution
Correct Option:
(4)
From the graph, the activation energy for the forward reaction is KJ. The change in enthalpy is given as KJ. The activation energy for the reverse reaction is:
Since KJ (because the products have lower energy than the reactants), we calculate:
30
PYQ 2025
medium
chemistryID: kcet-202
For the reaction , the initial concentration of is 2.0 mol L , and after 300 minutes, it is reduced to 1.4 mol L . The rate of production of (in mol L min ) is
1
2
3
4
Official Solution
Correct Option:
(3)
The rate of production of can be determined from the rate of change of . The balanced equation shows that for every 2 moles of consumed, 4 moles of are produced. Therefore, the rate of production of is: Using the concentration change of :
Since the reaction occurs over 300 minutes, the rate of change of is:
Thus, the rate of production of is:
31
PYQ 2026
medium
chemistryID: kcet-202
Given below are the half-cell reactions:
Mn + 2e Mn (E = -1.18 V)
Mn + e Mn (E = +1.51 V)
The E for 3 Mn Mn + 2Mn will be \rule{2cm{0.1mm}}
1
- 2.69 V, the reaction will not occur (Non-Spontaneous)
2
2.69 V, the reaction will occur (Spontaneous)
3
- 0.33 V, the reaction will not occur (Non-Spontaneous)
4
- 0.33 V, the reaction will occur (Spontaneous)
Official Solution
Correct Option:
(1)
Step 1: Identify the half-reactions as they occur in the overall reaction.
The overall reaction is: 3 Mn Mn + 2Mn .
Let's break this down into oxidation and reduction half-reactions:
Reduction: One Mn ion is gaining 2 electrons to become Mn.
Oxidation: Two Mn ions are losing electrons to become two Mn ions. This is the reverse of the second given half-reaction. When we reverse a half-reaction, we change the sign of its E . The given reduction is: Mn + e Mn (E = +1.51 V). So, the oxidation half-reaction is: Mn Mn + e (E = -1.51 V).
Step 2: Calculate the standard cell potential (E ).
The standard cell potential is the sum of the standard reduction potential and the standard oxidation potential.
Note: E is an intensive property, so we do not multiply it by the stoichiometric coefficients. We just add the potentials for the identified half-reactions.
Step 3: Determine the spontaneity of the reaction.
The spontaneity of an electrochemical reaction is determined by the sign of E .
If E >0, the reaction is spontaneous under standard conditions ( G <0).
If E <0, the reaction is non-spontaneous under standard conditions ( G >0).
Since E = -2.69 V, which is negative, the reaction is non-spontaneous and will not occur as written. Step 4: Final Answer.
The E is -2.69 V, and the reaction is non-spontaneous.
32
PYQ 2026
medium
chemistryID: kcet-202
The conductivity of centimolar solution of KCl at 298 K is 0.021 Ohm cm and the resistance of the cell containing the solution at 298 K is 60 . The value of cell constant (G*) is
1
3.28 cm
2
1.26 cm
3
3.34 cm
4
1.34 cm
Official Solution
Correct Option:
(2)
Step 1: Key Formula relating conductivity, resistance, and cell constant.
The relationship between these three quantities is:
The cell constant is defined as the ratio of the distance between the electrodes (l) to their area of cross-section (A), G* = l/A. Step 2: Identify the given values.
Conductivity ( ) = 0.021 cm
Resistance (R) = 60
Cell Constant (G*) = ?
The concentration ("centimolar", which is 0.01 M) is extra information and not needed for this specific calculation. Step 3: Rearrange the formula and calculate the cell constant.
From the formula , we can rearrange to solve for G*:
Substitute the given values:
The units are consistent: cm = cm . Step 4: Final Answer.
The value of the cell constant is 1.26 cm .
33
PYQ 2026
hard
chemistryID: kcet-202
For a reaction having three steps, the overall rate constant is . The values of , , and (activation energies stepwise) are 40, 50, and 60 kJ mol respectively. Then the overall (activation energy) of the reaction is .
1
30 kJ mol
2
40 kJ mol
3
50 kJ mol
4
60 kJ mol
Official Solution
Correct Option:
(3)
Step 1: Relate the overall rate constant to the Arrhenius equation.
The Arrhenius equation relates the rate constant (k) to the activation energy (E ):
We can apply this relationship to the overall rate constant K and the individual rate constants k , k , and k .
Step 2: Substitute the Arrhenius expressions into the given overall rate constant equation.
The given relation is:
Substituting the Arrhenius forms:
Assuming , we can focus on the exponential parts:
Using the rule and :
Step 3: Equate the exponents to find the overall activation energy (E ).
For the equality to hold, the exponents must be equal:
Multiplying by -RT, we get:
Step 4: Substitute the given values and calculate E .
We are given:
E = 40 kJ mol
E = 50 kJ mol
E = 60 kJ mol
Step 5: Final Answer.
The overall activation energy of the reaction is 50 kJ mol .
34
PYQ 2026
hard
chemistryID: kcet-202
Which one of the following graph is not applicable for a 1st order reaction (R P)?
1
A graph of [R] vs t with a downward sloping curve.
2
A graph of ln[R] vs t with a downward sloping straight line.
3
A graph of log [R] vs t with an upward sloping straight line.
4
A graph of log ([R] /[R]) vs t with an upward sloping straight line through the origin.
Official Solution
Correct Option:
(3)
Step 1: Key Formula - The Integrated Rate Law for a First-Order Reaction.
The integrated rate law for a first-order reaction can be expressed in several forms:
where [R] is the concentration of reactant at time t, [R] is the initial concentration, and k is the rate constant. Step 2: Analyze each graph based on the integrated rate laws. (A) [R] vs t: From equation (1), , the concentration of the reactant decreases exponentially with time. This corresponds to a downward sloping curve. This graph is applicable. (The image shows an upward arrow from [R], which is confusing, but the concept of a curve is correct). (B) ln[R] vs t: From equation (2), . This is in the form of a linear equation y = mx + c, where y = ln[R], x = t, the slope m = -k, and the y-intercept c = ln[R] . Since the slope (-k) is negative, this is a downward sloping straight line. This graph is applicable. (C) log [R] vs t: From equation (3), . This is also in the form y = mx + c, where y = log , x = t, and the slope m = . The slope is negative. The graph should be a downward sloping straight line. The option shows an upward sloping line. This graph is not applicable. (D) log ([R] /[R]) vs t: From equation (4), . This is in the form y = mx, where y = log([R] /[R]), x = t, and the slope m = . Since the slope is positive, this is an upward sloping straight line passing through the origin. This graph is applicable. Step 3: Final Answer.
The graph of log vs t should be a straight line with a negative slope. The graph shown in option (C) has a positive slope and is therefore not applicable to a first-order reaction.
35
PYQ 2026
medium
chemistryID: kcet-202
The activation energy for the reaction X Y is 150 kJ mol . The change in enthalpy for the above reaction is -135 kJ mol . Then the activation energy for Y X is _________.
1
280 kJ mol
2
285 kJ mol
3
270 kJ mol
4
15 kJ mol
Official Solution
Correct Option:
(2)
Step 1: Understanding the Question:
The question asks to determine the activation energy for the reverse reaction (Y X), given the activation energy for the forward reaction (X Y) and the enthalpy change for the forward reaction. Step 2: Key Formula or Approach:
The relationship between the activation energy of the forward reaction ( ), the activation energy of the reverse reaction ( ), and the enthalpy change of the reaction ( ) is:
This relationship can be visualized using a reaction coordinate diagram. For an exothermic reaction ( ), the products are at a lower energy level than the reactants. Step 3: Detailed Explanation:
Given values for the forward reaction :
Activation energy for forward reaction,
Change in enthalpy,
We need to find the activation energy for the reverse reaction , which is .
Substitute the given values into the formula:
Now, solve for :
Step 4: Final Answer:
The activation energy for the reverse reaction Y X is 285 kJ mol .
36
PYQ 2026
medium
chemistryID: kcet-202
Which one of the following graph is not applicable for a 1st order reaction (R P)?
1
Graph 1
2
Graph 2
3
Graph 3
4
Graph 4
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
The question asks to identify the graph that does NOT represent the kinetic behavior of a first-order reaction. Step 2: Key Formula or Approach:
For a first-order reaction , the integrated rate law is:
This can be rewritten in various forms for graphical analysis. Step 3: Detailed Explanation:
Let's analyze each type of graph for a first-order reaction:
1. Graph of vs : The integrated rate law is . This is in the form , where , (a negative slope), and . Therefore, a plot of vs is a straight line with a negative slope. This matches Graph (2) and is applicable. 2. Graph of vs : Dividing the integrated rate law by 2.303 gives . This is also a straight line with a negative slope. This matches Graph (3) and is applicable. 3. Graph of vs : From , we get . This is a straight line passing through the origin with a positive slope. This matches Graph (4) and is applicable. 4. Graph of vs : The concentration of reactant as a function of time is given by . This represents an exponential decay, meaning the concentration of reactant decreases over time, but not linearly. It is never a constant value unless (no reaction) or . Graph (1) shows as a horizontal line, implying constant concentration over time. This is incorrect for a reaction where reactant is consumed. Step 4: Final Answer:
Graph (1) is not applicable for a 1st order reaction.
37
PYQ 2026
medium
chemistryID: kcet-202
For a 1st order change R P, the concentration of Reactant R changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of R is 0.01 M is________.
1
1.73 10 M min
2
3.47 10 M min
3
3.47 10 M min
4
1.73 10 M min
Official Solution
Correct Option:
(2)
Step 1: Understanding the Question:
The question asks to calculate the instantaneous rate of a first-order reaction at a specific reactant concentration, given the initial and final concentrations over a time interval. Step 2: Key Formula or Approach:
1. Integrated Rate Law for First-Order Reaction:
Where: - is the rate constant. - is the time. - is the initial concentration of reactant. - is the concentration of reactant at time .
2. Rate Law for First-Order Reaction:
Step 3: Detailed Explanation: Part 1: Calculate the rate constant ( )
Given:
Initial concentration,
Concentration at time ,
Time,
Substitute these values into the integrated rate law:
Using :
Rounding to a reasonable number of significant figures, . Part 2: Calculate the rate of reaction when
Using the rate law for a first-order reaction:
Substitute the calculated and the given concentration:
Expressing in scientific notation:
Step 4: Final Answer:
The rate of reaction when the concentration of R is 0.01 M is M min .
38
PYQ 2026
medium
chemistryID: kcet-202
For a reaction having three steps, the overall rate constant is . The values E , E and E (activation energies stepwise) are 40, 50 and 60 kJ mol respectively. Then the overall E (activation energy) of the reaction is ______
1
30 kJ mol
2
40 kJ mol
3
50 kJ mol
4
60 kJ mol
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
The question asks to calculate the overall activation energy for a multi-step reaction, given the expression for its overall rate constant in terms of individual step rate constants and their respective activation energies. Step 2: Key Formula or Approach:
The Arrhenius equation relates the rate constant to the activation energy :
Taking the natural logarithm of both sides:
For an overall reaction, if its rate constant is expressed as a product or quotient of individual step rate constants, the overall activation energy can be found by summing and subtracting the individual activation energies. Step 3: Detailed Explanation:
Given the overall rate constant expression:
Take the natural logarithm of both sides:
Now, substitute the Arrhenius equation (in logarithmic form) for each rate constant:
Equating the terms containing the activation energies (which depend on ):
Multiply by to solve for the overall activation energy:
Given values:
Substitute these values:
Step 4: Final Answer:
The overall activation energy of the reaction is 30 kJ mol .
About Chemical Kinetics - KCET
Chemical Kinetics is a vital chapter for KCET aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
Frequently Asked Questions
Why focus on Chemical Kinetics PYQs?
Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within Chemical Kinetics carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
