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Q. No.For two chemical reactions A and B, if the difference between their activation energy is 20 KJ and the temperature is 300 K, then determine \(\ln \left(\frac{\mathrm{k}_2}{\mathrm{k}_1}\right)\):
[Use R= 8.3 J/mol-K]
1. 8.032
2. 4.016
3. 16.64
4. 2.303
[Use R= 8.3 J/mol-K]
1. 8.032
2. 4.016
3. 16.64
4. 2.303
Subtopic: Arrhenius Equation |
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Consider the reaction given below:
\(A_g \rightarrow B_g + C_g\)
The initial pressure of A is 1 bar after 100 minutes. The total pressure becomes 1.5 bar. Find the value of the rate constant of the reaction, assuming it as a first-order reaction:
1. \(6.9 \times 10^{-4} \text {min}^{-1}\)
2. \(6.9 \times 10^{-3} \text {min}^{-1}\)
3. \(39 \times 10^{-5} \text {min}^{-1}\)
4. \(3.9 \times 10^{-2} \text {min}^{-1}\)
\(A_g \rightarrow B_g + C_g\)
The initial pressure of A is 1 bar after 100 minutes. The total pressure becomes 1.5 bar. Find the value of the rate constant of the reaction, assuming it as a first-order reaction:
1. \(6.9 \times 10^{-4} \text {min}^{-1}\)
2. \(6.9 \times 10^{-3} \text {min}^{-1}\)
3. \(39 \times 10^{-5} \text {min}^{-1}\)
4. \(3.9 \times 10^{-2} \text {min}^{-1}\)
Subtopic: First Order Reaction Kinetics |
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The half-life of radioactive isotope Zn⁶⁵ is 245 days. Find the time after which the activity of a Zn⁶⁵ sample becomes 75% of its initial value.
Given:
Half-life (t₁/₂) = 245 days
Use:
log 4 = 0.6021
log 3 = 0.4771
1. 96
2. 100
3. 102
4. 108
Subtopic: First Order Reaction Kinetics |
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Level 3: 35%-60%
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For a reaction at 300 K, on addition of catalyst, activation energy of reaction lowered by 10 kJ.
Then calculate the value of \(\log \frac{\mathrm{K}_{\text {catalysed }}}{\mathrm{K}_{\text {uncatalysed }}}\):
1. 1.74
2. 0.174
3. 17.4
4. 3.48
Then calculate the value of \(\log \frac{\mathrm{K}_{\text {catalysed }}}{\mathrm{K}_{\text {uncatalysed }}}\):
1. 1.74
2. 0.174
3. 17.4
4. 3.48
Subtopic: Arrhenius Equation |
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Level 2: 60%+
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For a chemical reaction: \(A\rightarrow D\)
Mechanism is expressed in three steps:
Step -1 : \(\mathrm{A} \rightarrow \mathrm{~B}: \Delta \mathrm{H}=+\mathrm{ve}\)
Step -2 : \( B \rightarrow C: \Delta H=-\mathrm{ve} \)
Step -3 : \( C \rightarrow D: \Delta H=-\mathrm{ve}\)
1.
2. 
3.
4. 
Mechanism is expressed in three steps:
Step -1 : \(\mathrm{A} \rightarrow \mathrm{~B}: \Delta \mathrm{H}=+\mathrm{ve}\)
Step -2 : \( B \rightarrow C: \Delta H=-\mathrm{ve} \)
Step -3 : \( C \rightarrow D: \Delta H=-\mathrm{ve}\)
1.
2. 3.
4. Subtopic: First Order Reaction Kinetics | Order, Molecularity and Mechanism |
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Given below are two reactions with their activation energies:
\(\begin{aligned} & \mathrm{A} \rightarrow \mathrm{~B} ; \mathrm{E}_{\mathrm{a}_1} \\ & \mathrm{C} \rightarrow \mathrm{D} ; \mathrm{E}_{\mathrm{a}_2} \end{aligned}\)
\(\log _{10} \mathrm{K} \) for first reaction \(=14.34-\dfrac{1.5 \times 10^4}{\mathrm{~T}}\)
\(E_{a_2}\) is \(1 / 5^{\text {th }}\) of \(E_{a_1}.\) Then the value of \(E_{a_2}\) (in kJ/mol) is :
1. 38
2. 54
3. 27
4. 80
\(\begin{aligned} & \mathrm{A} \rightarrow \mathrm{~B} ; \mathrm{E}_{\mathrm{a}_1} \\ & \mathrm{C} \rightarrow \mathrm{D} ; \mathrm{E}_{\mathrm{a}_2} \end{aligned}\)
\(\log _{10} \mathrm{K} \) for first reaction \(=14.34-\dfrac{1.5 \times 10^4}{\mathrm{~T}}\)
\(E_{a_2}\) is \(1 / 5^{\text {th }}\) of \(E_{a_1}.\) Then the value of \(E_{a_2}\) (in kJ/mol) is :
1. 38
2. 54
3. 27
4. 80
Subtopic: First Order Reaction Kinetics | Arrhenius Equation |
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Consider the following statement(s) about Arrhenius equation:
(A) The fraction of particles having energy less than activation energy is \(\mathrm{e}^{-\dfrac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}}~\).
(B) Reaction with lower activation energy is faster.
(C) On increasing temperature by \(10^oC\), rate of reaction doubles.
(D) Graph of log K v/s \(\frac1 T\) is a straight line with slope \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}} .\)
Select correct statement:
1. A and B are correct
2. B and D are correct
3. B and C are correct
4. C and D are correct
(A) The fraction of particles having energy less than activation energy is \(\mathrm{e}^{-\dfrac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}}~\).
(B) Reaction with lower activation energy is faster.
(C) On increasing temperature by \(10^oC\), rate of reaction doubles.
(D) Graph of log K v/s \(\frac1 T\) is a straight line with slope \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{R}} .\)
Select correct statement:
1. A and B are correct
2. B and D are correct
3. B and C are correct
4. C and D are correct
Subtopic: Definition, Rate Constant, Rate Law | Arrhenius Equation |
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Three experiments are running in separate vessel, following \(1^{st}\) order kinetics:
Experiment-(A) 100 ml, 10 M
Experiment-(B) 200 ml, 10 M
Experiment-(C) 100 ml, 10 M + 100 ml \(H_2O\)
Select correct order of rate of reaction in above experiments:
1. A = B = C
2. A = B > C
3. A > B > C
4. C > A > B
Experiment-(A) 100 ml, 10 M
Experiment-(B) 200 ml, 10 M
Experiment-(C) 100 ml, 10 M + 100 ml \(H_2O\)
Select correct order of rate of reaction in above experiments:
1. A = B = C
2. A = B > C
3. A > B > C
4. C > A > B
Subtopic: First Order Reaction Kinetics |
Level 3: 35%-60%
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For a given reaction \(\mathrm{A} \rightarrow \mathrm{nB}\), a graph is given between concentration and time. Find value of n for above reaction, based on the information given in graph for 10 min.
1. 5
2. 3
3. 6
4. 8
1. 5
2. 3
3. 6
4. 8
Subtopic: Definition, Rate Constant, Rate Law |
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Level 1: 80%+
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Given at 10 AM, reaction is started
(i) \(\mathrm{A} \xrightarrow{\mathrm{k}}\) Product (\(1^{st}\) order reaction)
(ii) \(\mathrm{BrO}_3^{-}+5 \mathrm{Br}^{-} \rightarrow 3 \mathrm{Br}_2\)
At 10:10 AM, rate of disappearance of \(Br^–\) was \(2 × 10^{–3}\) M/min. and concentration of A was 0.1 M, if both reactions were proceed with same rate at this time then value of k will be ?
1. \( 10^{-3} \mathrm{~min}^{-1}\)
2. \( 2 \times 10^{-3} \mathrm{~min}^{-1}\)
3. \( 4 \times 10^{-3} \mathrm{~min}^{-1}\)
4. \( 8 \times 10^{-3} \mathrm{~min}^{-1}\)
(i) \(\mathrm{A} \xrightarrow{\mathrm{k}}\) Product (\(1^{st}\) order reaction)
(ii) \(\mathrm{BrO}_3^{-}+5 \mathrm{Br}^{-} \rightarrow 3 \mathrm{Br}_2\)
At 10:10 AM, rate of disappearance of \(Br^–\) was \(2 × 10^{–3}\) M/min. and concentration of A was 0.1 M, if both reactions were proceed with same rate at this time then value of k will be ?
1. \( 10^{-3} \mathrm{~min}^{-1}\)
2. \( 2 \times 10^{-3} \mathrm{~min}^{-1}\)
3. \( 4 \times 10^{-3} \mathrm{~min}^{-1}\)
4. \( 8 \times 10^{-3} \mathrm{~min}^{-1}\)
Subtopic: First Order Reaction Kinetics |
70%
Level 2: 60%+
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