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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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 |
Level 4: Below 35%
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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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Consider the following graph:
What is the correct order of increasing activation energies \(\mathrm{(E_a)}\)?
1. \(\mathrm {E_{a_2}>E_{a_1}>E_{a_3}}\)
2. \(\mathrm{E_{a_1}>E_{a_2}>E_{a_3}}\)
3. \(\mathrm{E_{a_3}>E_{a_2}>E_{a_1}} \)
4. \(\mathrm{E_{a_2}>E_{a_3}>E_{a_1}}\)
What is the correct order of increasing activation energies \(\mathrm{(E_a)}\)?
1. \(\mathrm {E_{a_2}>E_{a_1}>E_{a_3}}\)
2. \(\mathrm{E_{a_1}>E_{a_2}>E_{a_3}}\)
3. \(\mathrm{E_{a_3}>E_{a_2}>E_{a_1}} \)
4. \(\mathrm{E_{a_2}>E_{a_3}>E_{a_1}}\)
Subtopic: Arrhenius Equation |
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Which of the following statement is correct w.r.t. Arrhenius equation?
1. Dimensions of \(\mathrm{k}\) and \(\mathrm{A}\) are same
2. \(\mathrm{k}\) decreases with increase in temperature generally
3. \(\mathrm{A}\) decreases with increase in temperature always
4. \(\mathrm{k}\) increases as value of \(E_a\) increase
1. Dimensions of \(\mathrm{k}\) and \(\mathrm{A}\) are same
2. \(\mathrm{k}\) decreases with increase in temperature generally
3. \(\mathrm{A}\) decreases with increase in temperature always
4. \(\mathrm{k}\) increases as value of \(E_a\) increase
Subtopic: Arrhenius Equation |
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Consider the following reaction sequence :
Overall \(k = {k_1k_2 \over k_3}\)
if \(E_{a_1} = 300 kJ/mole \\ E_{a_2} = 200 kJ/mole \)
overall \((E_a)_{eff} = 400 kJ/mole \)
Find out \(E_{a_3} \) (in kJ/mole):
1. 200
2. 400
3. 100
4. 900
\(\)
Overall \(k = {k_1k_2 \over k_3}\)
if \(E_{a_1} = 300 kJ/mole \\ E_{a_2} = 200 kJ/mole \)
overall \((E_a)_{eff} = 400 kJ/mole \)
Find out \(E_{a_3} \) (in kJ/mole):
1. 200
2. 400
3. 100
4. 900
\(\)
Subtopic: Arrhenius Equation |
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Given, \(\mathrm{K}_{\text {net }}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3} \text { when } \mathrm{E}_{\mathrm{a}_1}=40 \mathrm{~kJ} / \mathrm{mol}\),
\(\mathrm{E}_{\mathrm{a}_2}=50 \mathrm{~kJ} / \mathrm{mol} ~ \text {and}~\mathrm{E}_{\mathrm{a}_3}=60 \mathrm{~kJ} / \mathrm{mol} \text {}\)
Calculate value of \(\left(\mathrm{E}_{\mathrm{a}}\right)_{\text {net }} \text { in } \mathrm{kJ} / \mathrm{mol}\):
1. 90
2. 30
3. 70
4. 40
\(\mathrm{E}_{\mathrm{a}_2}=50 \mathrm{~kJ} / \mathrm{mol} ~ \text {and}~\mathrm{E}_{\mathrm{a}_3}=60 \mathrm{~kJ} / \mathrm{mol} \text {}\)
Calculate value of \(\left(\mathrm{E}_{\mathrm{a}}\right)_{\text {net }} \text { in } \mathrm{kJ} / \mathrm{mol}\):
1. 90
2. 30
3. 70
4. 40
Subtopic: Arrhenius Equation |
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For a 1st order reaction, following graph is obtained between lnk and \({1000 \over T}\). Then, the activation energy of the reaction in kcal is:
1. 37 kcal
2. 40 kcal
3. 42 kcal
4. 34 kcal
1. 37 kcal
2. 40 kcal
3. 42 kcal
4. 34 kcal
Subtopic: Arrhenius Equation |
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Consider the equilibrium reaction: A(g) ⇌ B(g) with an enthalpy change (ΔH) of -42 kJ/mol.
Determine the activation energies for the forward and backward reactions, given that the ratio of the activation energy of the forward reaction to the activation energy of the backward reaction is 2 : 3.
| 1. | 84 kJ/mole, 126 kJ/mole | 2. | 24 kJ/mole, 36 kJ/mole |
| 3. | 48 kJ/mole, 72 kJ/mole | 4. | 90 kJ/mole, 135 kJ/mole |
Subtopic: Arrhenius Equation |
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