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CHXII04:CHEMICAL KINETICS

320209 The maximum value of activation energy is equal to:

1 Zero

2 Heat of the reaction

3 Threshold energy

4 None of these

CHXII04:CHEMICAL KINETICS

320210 The Arrhenius equation for trans isomerisation of 2-butene and 1-butene nitrile are given as follows:

(i) For 2-butene; $k (s^{- 1}) = 1 0^{13 .8} e^{(- 263 .5 k J / m o l / / R T)}$

(ii) For 2-butene nitrile ; $k' (s^{- 1}) = 1 0^{11} ex p^{- 214 .5kmo \Gamma^{- 1/RT}}$
The temperature at which $k = k'$ is

1

913.87 K

2

533.43 K

3

1000.02 K

4

407.05 K

Explanation:

$10^{13.8} \exp^{- 263.5 k J m o Γ^{1 / R T}} = 10^{11} \exp^{- 214.5 k J m o I^{- 1 / R T}}$
or
$\begin{matrix} 10^{13.8 - 11} = \exp \frac{[(- 214.5) - (- 263.5)] {kJmol}^{- 1}}{8.314 \times 10^{- 3} {kJmol}^{- 1} K^{- 1} \times T} \\ or 630.957 = \exp \frac{5.893 \times 10^{3}}{T} \\ ∴ 2.303 \log 630.957 = \frac{5.893 \times 10^{3}}{T} \\ ∴ T = \frac{5.893 \times 10^{3}}{2.303 \times 2.8} = 913.87 K \end{matrix}$

CHXII04:CHEMICAL KINETICS

320211 At room temperature, the reaction between NO and $O_{2}$ to give ${NO}_{2}$ is fast, while that between CO and $O_{2}$ is slow. It is due to

1 CO is smaller in size than NO

2 CO is poisonous

3 The activation energy for the reaction,

2 NO + O_{2} \to 2 {NO}_{2}

is less than

2 CO + O_{2} \to 2 {CO}_{2}

4 None of the above

CHXII04:CHEMICAL KINETICS

320212 The rate of a reaction A doubles on increasing the temperature from 300 to $310 K$ . By how much, the temperature of reaction $B$ should be increased from $300 K$ so that rate doubles if activation energy of the reaction $B$ is twice that of reaction $A$ .

1

4.92 K

2

9.84 K

3

19.67 K

4

2.45 K

Explanation:

$\ln (\frac{k_{1}}{k_{2}}) = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$l n (2) = \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} (1)$
$l n (2) = 2 \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{T}} (2)$
From (1) and (2)
$\frac{2 E_{a}}{R} {\frac{1}{300} - \frac{1}{T}}$
$= \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} \frac{1}{300} + \frac{1}{310} = \frac{2}{T}$
$\Rightarrow T = \frac{300 \times 310}{610} \times 2$
$T = 304 .92$
Incrase in $T = 304.92 - 300 = 4.92 K$

JEE - 2017

CHXII04:CHEMICAL KINETICS

320213 Consider the following reaction,
supporting img
The reaction is of first order in each diagram, with an equilibrium constant of $10^{4}$ . For the conversion of chair form to boat form $e^{- E/aRT} = 4.35 \times 10^{- 8}$ at 298 K with pre-exponential factor of $10^{12} s^{- 1} .$ Apparent rate constant $(= k_{A} / k_{B})$ at 298 K is

1

4.35 \times 10^{4} s^{- 1}

2

4.35 \times 10^{8} s^{- 1}

3

4.35 \times 10^{- 8} s^{- 1}

4

4.35 \times 10^{12} s^{- 1}

Explanation:

$k_{B} = A e^{- E_{a} /RT}$
$= 10^{12} \times 4.35 \times 10^{- 8} = 4.35 \times 10^{4} s^{- 1}$
Also equilibrium constant, $K = \frac{k_{A}}{k_{B}} = 10^{4}$
$∴ k_{A} = k_{B} \times 10^{4} = 4.35 \times 10^{8} s^{- 1}$

CHXII04:CHEMICAL KINETICS

320209 The maximum value of activation energy is equal to:

1 Zero

2 Heat of the reaction

3 Threshold energy

4 None of these

CHXII04:CHEMICAL KINETICS

320210 The Arrhenius equation for trans isomerisation of 2-butene and 1-butene nitrile are given as follows:

(i) For 2-butene; $k (s^{- 1}) = 1 0^{13 .8} e^{(- 263 .5 k J / m o l / / R T)}$

(ii) For 2-butene nitrile ; $k' (s^{- 1}) = 1 0^{11} ex p^{- 214 .5kmo \Gamma^{- 1/RT}}$
The temperature at which $k = k'$ is

1

913.87 K

2

533.43 K

3

1000.02 K

4

407.05 K

Explanation:

$10^{13.8} \exp^{- 263.5 k J m o Γ^{1 / R T}} = 10^{11} \exp^{- 214.5 k J m o I^{- 1 / R T}}$
or
$\begin{matrix} 10^{13.8 - 11} = \exp \frac{[(- 214.5) - (- 263.5)] {kJmol}^{- 1}}{8.314 \times 10^{- 3} {kJmol}^{- 1} K^{- 1} \times T} \\ or 630.957 = \exp \frac{5.893 \times 10^{3}}{T} \\ ∴ 2.303 \log 630.957 = \frac{5.893 \times 10^{3}}{T} \\ ∴ T = \frac{5.893 \times 10^{3}}{2.303 \times 2.8} = 913.87 K \end{matrix}$

CHXII04:CHEMICAL KINETICS

320211 At room temperature, the reaction between NO and $O_{2}$ to give ${NO}_{2}$ is fast, while that between CO and $O_{2}$ is slow. It is due to

1 CO is smaller in size than NO

2 CO is poisonous

3 The activation energy for the reaction,

2 NO + O_{2} \to 2 {NO}_{2}

is less than

2 CO + O_{2} \to 2 {CO}_{2}

4 None of the above

CHXII04:CHEMICAL KINETICS

320212 The rate of a reaction A doubles on increasing the temperature from 300 to $310 K$ . By how much, the temperature of reaction $B$ should be increased from $300 K$ so that rate doubles if activation energy of the reaction $B$ is twice that of reaction $A$ .

1

4.92 K

2

9.84 K

3

19.67 K

4

2.45 K

Explanation:

$\ln (\frac{k_{1}}{k_{2}}) = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$l n (2) = \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} (1)$
$l n (2) = 2 \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{T}} (2)$
From (1) and (2)
$\frac{2 E_{a}}{R} {\frac{1}{300} - \frac{1}{T}}$
$= \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} \frac{1}{300} + \frac{1}{310} = \frac{2}{T}$
$\Rightarrow T = \frac{300 \times 310}{610} \times 2$
$T = 304 .92$
Incrase in $T = 304.92 - 300 = 4.92 K$

JEE - 2017

CHXII04:CHEMICAL KINETICS

320213 Consider the following reaction,
supporting img
The reaction is of first order in each diagram, with an equilibrium constant of $10^{4}$ . For the conversion of chair form to boat form $e^{- E/aRT} = 4.35 \times 10^{- 8}$ at 298 K with pre-exponential factor of $10^{12} s^{- 1} .$ Apparent rate constant $(= k_{A} / k_{B})$ at 298 K is

1

4.35 \times 10^{4} s^{- 1}

2

4.35 \times 10^{8} s^{- 1}

3

4.35 \times 10^{- 8} s^{- 1}

4

4.35 \times 10^{12} s^{- 1}

Explanation:

$k_{B} = A e^{- E_{a} /RT}$
$= 10^{12} \times 4.35 \times 10^{- 8} = 4.35 \times 10^{4} s^{- 1}$
Also equilibrium constant, $K = \frac{k_{A}}{k_{B}} = 10^{4}$
$∴ k_{A} = k_{B} \times 10^{4} = 4.35 \times 10^{8} s^{- 1}$

CHXII04:CHEMICAL KINETICS

320209 The maximum value of activation energy is equal to:

1 Zero

2 Heat of the reaction

3 Threshold energy

4 None of these

CHXII04:CHEMICAL KINETICS

320210 The Arrhenius equation for trans isomerisation of 2-butene and 1-butene nitrile are given as follows:

(i) For 2-butene; $k (s^{- 1}) = 1 0^{13 .8} e^{(- 263 .5 k J / m o l / / R T)}$

(ii) For 2-butene nitrile ; $k' (s^{- 1}) = 1 0^{11} ex p^{- 214 .5kmo \Gamma^{- 1/RT}}$
The temperature at which $k = k'$ is

1

913.87 K

2

533.43 K

3

1000.02 K

4

407.05 K

Explanation:

$10^{13.8} \exp^{- 263.5 k J m o Γ^{1 / R T}} = 10^{11} \exp^{- 214.5 k J m o I^{- 1 / R T}}$
or
$\begin{matrix} 10^{13.8 - 11} = \exp \frac{[(- 214.5) - (- 263.5)] {kJmol}^{- 1}}{8.314 \times 10^{- 3} {kJmol}^{- 1} K^{- 1} \times T} \\ or 630.957 = \exp \frac{5.893 \times 10^{3}}{T} \\ ∴ 2.303 \log 630.957 = \frac{5.893 \times 10^{3}}{T} \\ ∴ T = \frac{5.893 \times 10^{3}}{2.303 \times 2.8} = 913.87 K \end{matrix}$

CHXII04:CHEMICAL KINETICS

320211 At room temperature, the reaction between NO and $O_{2}$ to give ${NO}_{2}$ is fast, while that between CO and $O_{2}$ is slow. It is due to

1 CO is smaller in size than NO

2 CO is poisonous

3 The activation energy for the reaction,

2 NO + O_{2} \to 2 {NO}_{2}

is less than

2 CO + O_{2} \to 2 {CO}_{2}

4 None of the above

CHXII04:CHEMICAL KINETICS

320212 The rate of a reaction A doubles on increasing the temperature from 300 to $310 K$ . By how much, the temperature of reaction $B$ should be increased from $300 K$ so that rate doubles if activation energy of the reaction $B$ is twice that of reaction $A$ .

1

4.92 K

2

9.84 K

3

19.67 K

4

2.45 K

Explanation:

$\ln (\frac{k_{1}}{k_{2}}) = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$l n (2) = \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} (1)$
$l n (2) = 2 \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{T}} (2)$
From (1) and (2)
$\frac{2 E_{a}}{R} {\frac{1}{300} - \frac{1}{T}}$
$= \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} \frac{1}{300} + \frac{1}{310} = \frac{2}{T}$
$\Rightarrow T = \frac{300 \times 310}{610} \times 2$
$T = 304 .92$
Incrase in $T = 304.92 - 300 = 4.92 K$

JEE - 2017

CHXII04:CHEMICAL KINETICS

320213 Consider the following reaction,
supporting img
The reaction is of first order in each diagram, with an equilibrium constant of $10^{4}$ . For the conversion of chair form to boat form $e^{- E/aRT} = 4.35 \times 10^{- 8}$ at 298 K with pre-exponential factor of $10^{12} s^{- 1} .$ Apparent rate constant $(= k_{A} / k_{B})$ at 298 K is

1

4.35 \times 10^{4} s^{- 1}

2

4.35 \times 10^{8} s^{- 1}

3

4.35 \times 10^{- 8} s^{- 1}

4

4.35 \times 10^{12} s^{- 1}

Explanation:

$k_{B} = A e^{- E_{a} /RT}$
$= 10^{12} \times 4.35 \times 10^{- 8} = 4.35 \times 10^{4} s^{- 1}$
Also equilibrium constant, $K = \frac{k_{A}}{k_{B}} = 10^{4}$
$∴ k_{A} = k_{B} \times 10^{4} = 4.35 \times 10^{8} s^{- 1}$

CHXII04:CHEMICAL KINETICS

320209 The maximum value of activation energy is equal to:

1 Zero

2 Heat of the reaction

3 Threshold energy

4 None of these

CHXII04:CHEMICAL KINETICS

320210 The Arrhenius equation for trans isomerisation of 2-butene and 1-butene nitrile are given as follows:

(i) For 2-butene; $k (s^{- 1}) = 1 0^{13 .8} e^{(- 263 .5 k J / m o l / / R T)}$

(ii) For 2-butene nitrile ; $k' (s^{- 1}) = 1 0^{11} ex p^{- 214 .5kmo \Gamma^{- 1/RT}}$
The temperature at which $k = k'$ is

1

913.87 K

2

533.43 K

3

1000.02 K

4

407.05 K

Explanation:

$10^{13.8} \exp^{- 263.5 k J m o Γ^{1 / R T}} = 10^{11} \exp^{- 214.5 k J m o I^{- 1 / R T}}$
or
$\begin{matrix} 10^{13.8 - 11} = \exp \frac{[(- 214.5) - (- 263.5)] {kJmol}^{- 1}}{8.314 \times 10^{- 3} {kJmol}^{- 1} K^{- 1} \times T} \\ or 630.957 = \exp \frac{5.893 \times 10^{3}}{T} \\ ∴ 2.303 \log 630.957 = \frac{5.893 \times 10^{3}}{T} \\ ∴ T = \frac{5.893 \times 10^{3}}{2.303 \times 2.8} = 913.87 K \end{matrix}$

CHXII04:CHEMICAL KINETICS

320211 At room temperature, the reaction between NO and $O_{2}$ to give ${NO}_{2}$ is fast, while that between CO and $O_{2}$ is slow. It is due to

1 CO is smaller in size than NO

2 CO is poisonous

3 The activation energy for the reaction,

2 NO + O_{2} \to 2 {NO}_{2}

is less than

2 CO + O_{2} \to 2 {CO}_{2}

4 None of the above

CHXII04:CHEMICAL KINETICS

320212 The rate of a reaction A doubles on increasing the temperature from 300 to $310 K$ . By how much, the temperature of reaction $B$ should be increased from $300 K$ so that rate doubles if activation energy of the reaction $B$ is twice that of reaction $A$ .

1

4.92 K

2

9.84 K

3

19.67 K

4

2.45 K

Explanation:

$\ln (\frac{k_{1}}{k_{2}}) = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$l n (2) = \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} (1)$
$l n (2) = 2 \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{T}} (2)$
From (1) and (2)
$\frac{2 E_{a}}{R} {\frac{1}{300} - \frac{1}{T}}$
$= \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} \frac{1}{300} + \frac{1}{310} = \frac{2}{T}$
$\Rightarrow T = \frac{300 \times 310}{610} \times 2$
$T = 304 .92$
Incrase in $T = 304.92 - 300 = 4.92 K$

JEE - 2017

CHXII04:CHEMICAL KINETICS

320213 Consider the following reaction,
supporting img
The reaction is of first order in each diagram, with an equilibrium constant of $10^{4}$ . For the conversion of chair form to boat form $e^{- E/aRT} = 4.35 \times 10^{- 8}$ at 298 K with pre-exponential factor of $10^{12} s^{- 1} .$ Apparent rate constant $(= k_{A} / k_{B})$ at 298 K is

1

4.35 \times 10^{4} s^{- 1}

2

4.35 \times 10^{8} s^{- 1}

3

4.35 \times 10^{- 8} s^{- 1}

4

4.35 \times 10^{12} s^{- 1}

Explanation:

$k_{B} = A e^{- E_{a} /RT}$
$= 10^{12} \times 4.35 \times 10^{- 8} = 4.35 \times 10^{4} s^{- 1}$
Also equilibrium constant, $K = \frac{k_{A}}{k_{B}} = 10^{4}$
$∴ k_{A} = k_{B} \times 10^{4} = 4.35 \times 10^{8} s^{- 1}$

CHXII04:CHEMICAL KINETICS

320209 The maximum value of activation energy is equal to:

1 Zero

2 Heat of the reaction

3 Threshold energy

4 None of these

CHXII04:CHEMICAL KINETICS

320210 The Arrhenius equation for trans isomerisation of 2-butene and 1-butene nitrile are given as follows:

(i) For 2-butene; $k (s^{- 1}) = 1 0^{13 .8} e^{(- 263 .5 k J / m o l / / R T)}$

(ii) For 2-butene nitrile ; $k' (s^{- 1}) = 1 0^{11} ex p^{- 214 .5kmo \Gamma^{- 1/RT}}$
The temperature at which $k = k'$ is

1

913.87 K

2

533.43 K

3

1000.02 K

4

407.05 K

Explanation:

$10^{13.8} \exp^{- 263.5 k J m o Γ^{1 / R T}} = 10^{11} \exp^{- 214.5 k J m o I^{- 1 / R T}}$
or
$\begin{matrix} 10^{13.8 - 11} = \exp \frac{[(- 214.5) - (- 263.5)] {kJmol}^{- 1}}{8.314 \times 10^{- 3} {kJmol}^{- 1} K^{- 1} \times T} \\ or 630.957 = \exp \frac{5.893 \times 10^{3}}{T} \\ ∴ 2.303 \log 630.957 = \frac{5.893 \times 10^{3}}{T} \\ ∴ T = \frac{5.893 \times 10^{3}}{2.303 \times 2.8} = 913.87 K \end{matrix}$

CHXII04:CHEMICAL KINETICS

320211 At room temperature, the reaction between NO and $O_{2}$ to give ${NO}_{2}$ is fast, while that between CO and $O_{2}$ is slow. It is due to

1 CO is smaller in size than NO

2 CO is poisonous

3 The activation energy for the reaction,

2 NO + O_{2} \to 2 {NO}_{2}

is less than

2 CO + O_{2} \to 2 {CO}_{2}

4 None of the above

CHXII04:CHEMICAL KINETICS

320212 The rate of a reaction A doubles on increasing the temperature from 300 to $310 K$ . By how much, the temperature of reaction $B$ should be increased from $300 K$ so that rate doubles if activation energy of the reaction $B$ is twice that of reaction $A$ .

1

4.92 K

2

9.84 K

3

19.67 K

4

2.45 K

Explanation:

$\ln (\frac{k_{1}}{k_{2}}) = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$l n (2) = \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} (1)$
$l n (2) = 2 \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{T}} (2)$
From (1) and (2)
$\frac{2 E_{a}}{R} {\frac{1}{300} - \frac{1}{T}}$
$= \frac{E_{a}}{R} {\frac{1}{300} - \frac{1}{310}} \frac{1}{300} + \frac{1}{310} = \frac{2}{T}$
$\Rightarrow T = \frac{300 \times 310}{610} \times 2$
$T = 304 .92$
Incrase in $T = 304.92 - 300 = 4.92 K$

JEE - 2017

CHXII04:CHEMICAL KINETICS

320213 Consider the following reaction,
supporting img
The reaction is of first order in each diagram, with an equilibrium constant of $10^{4}$ . For the conversion of chair form to boat form $e^{- E/aRT} = 4.35 \times 10^{- 8}$ at 298 K with pre-exponential factor of $10^{12} s^{- 1} .$ Apparent rate constant $(= k_{A} / k_{B})$ at 298 K is

1

4.35 \times 10^{4} s^{- 1}

2

4.35 \times 10^{8} s^{- 1}

3

4.35 \times 10^{- 8} s^{- 1}

4

4.35 \times 10^{12} s^{- 1}

Explanation:

$k_{B} = A e^{- E_{a} /RT}$
$= 10^{12} \times 4.35 \times 10^{- 8} = 4.35 \times 10^{4} s^{- 1}$
Also equilibrium constant, $K = \frac{k_{A}}{k_{B}} = 10^{4}$
$∴ k_{A} = k_{B} \times 10^{4} = 4.35 \times 10^{8} s^{- 1}$

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