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CHXII04:CHEMICAL KINETICS

320252 The activation energy for a simple chemical
reaction $A \to B$ is $E_{a}$ in forward direction. The
activation energy for reverse reaction

1 is always less than

E_{a}

2 can be less than or more than

E_{a}

3 is always double of

E_{a}

4 is negative of

E_{a}

Explanation:

Since the nature of reaction (i.e. exothermic or endothermic) is not given, $E_{a}$ for reverse reaction can be more or less.

CHXII04:CHEMICAL KINETICS

320253 The decomposition of $N_{2} O into N_{2}$ and in presence gaseous argon follow second order kinetics with $k = (5 .0 \times 1 0^{11} L m o l^{- 1} s^{- 1}) e^{- \frac{41570}{T}} K$ stands for Kelvin units). The energy of activation of the reaction is

1

5.0 \times 10^{11} J

2

41570 J

3

5000 J

4

345612.98 J

Explanation:

$E_{a} = 41570 \times R = 41570 \times 8 .31 = 345612 .98 J$

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320254 For a reaction, the value of rate constant at 300 $K$ is $6.0 \times 10^{5} s^{- 1}$ . The value of Arrhenius factor $A$ , at infinitely high temperature is:

1

\frac{6 \times 10^{- 5}}{300}

2

6 \times 10^{5}

3

6 \times 10^{5} \times e^{- Ea / 300 R}

4

e^{- Ea / 300 R}

Explanation:

$k = A e^{\frac{- Ea}{RT}}$
At $T = \infty; k = A$
$= 6 \times 10^{5} s^{- 1}$

KCET - 2023

CHXII04:CHEMICAL KINETICS

320255 A plot of $\ln k$ versus $\frac{1}{T}$ for a reaction gives the slope $- 2.0 \times 10^{3} K$ . The energy of activation for the reaction is ____ $kJ {mol}^{- 1}$ .

1 16.63

2 20.75

3 26.63

4 8.31

Explanation:

In a plot of $\ln k vs \frac{1}{T}$ ,
$-$ slope $= \frac{E_{a}}{R}$
$2 \times 10^{3} = \frac{E_{a}}{8.314}$
$E_{a} = 8.314 \times 2 \times 10^{3} = 16628 J = 16.63 kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320256 If for two reactions $E_{a_{1}} > E_{a_{2}}$ and ${TC}_{1}$ and ${TC}_{2}$ are temperature coefficient respectively, then which alternative is correct?

1

{TC}_{1} > {TC}_{2}

2

{TC}_{1} < {TC}_{2}

3

{TC}_{1} = {TC}_{2}

4 cannot be determined

Explanation:

We, know,
$l n (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$\frac{k_{2}}{k_{1}} = e^{\frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{1} = e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{2} = e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$\frac{T C_{1}}{T C_{2}} = \frac{e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}}{e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}} = e^{\frac{{(E_{a})}_{1} - {(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
If ${(E_{a})}_{1} > {(E_{a})}_{2}, t h e n T C_{1} < T C_{2}$ .

CHXII04:CHEMICAL KINETICS

320252 The activation energy for a simple chemical
reaction $A \to B$ is $E_{a}$ in forward direction. The
activation energy for reverse reaction

1 is always less than

E_{a}

2 can be less than or more than

E_{a}

3 is always double of

E_{a}

4 is negative of

E_{a}

Explanation:

Since the nature of reaction (i.e. exothermic or endothermic) is not given, $E_{a}$ for reverse reaction can be more or less.

CHXII04:CHEMICAL KINETICS

320253 The decomposition of $N_{2} O into N_{2}$ and in presence gaseous argon follow second order kinetics with $k = (5 .0 \times 1 0^{11} L m o l^{- 1} s^{- 1}) e^{- \frac{41570}{T}} K$ stands for Kelvin units). The energy of activation of the reaction is

1

5.0 \times 10^{11} J

2

41570 J

3

5000 J

4

345612.98 J

Explanation:

$E_{a} = 41570 \times R = 41570 \times 8 .31 = 345612 .98 J$

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320254 For a reaction, the value of rate constant at 300 $K$ is $6.0 \times 10^{5} s^{- 1}$ . The value of Arrhenius factor $A$ , at infinitely high temperature is:

1

\frac{6 \times 10^{- 5}}{300}

2

6 \times 10^{5}

3

6 \times 10^{5} \times e^{- Ea / 300 R}

4

e^{- Ea / 300 R}

Explanation:

$k = A e^{\frac{- Ea}{RT}}$
At $T = \infty; k = A$
$= 6 \times 10^{5} s^{- 1}$

KCET - 2023

CHXII04:CHEMICAL KINETICS

320255 A plot of $\ln k$ versus $\frac{1}{T}$ for a reaction gives the slope $- 2.0 \times 10^{3} K$ . The energy of activation for the reaction is ____ $kJ {mol}^{- 1}$ .

1 16.63

2 20.75

3 26.63

4 8.31

Explanation:

In a plot of $\ln k vs \frac{1}{T}$ ,
$-$ slope $= \frac{E_{a}}{R}$
$2 \times 10^{3} = \frac{E_{a}}{8.314}$
$E_{a} = 8.314 \times 2 \times 10^{3} = 16628 J = 16.63 kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320256 If for two reactions $E_{a_{1}} > E_{a_{2}}$ and ${TC}_{1}$ and ${TC}_{2}$ are temperature coefficient respectively, then which alternative is correct?

1

{TC}_{1} > {TC}_{2}

2

{TC}_{1} < {TC}_{2}

3

{TC}_{1} = {TC}_{2}

4 cannot be determined

Explanation:

We, know,
$l n (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$\frac{k_{2}}{k_{1}} = e^{\frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{1} = e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{2} = e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$\frac{T C_{1}}{T C_{2}} = \frac{e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}}{e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}} = e^{\frac{{(E_{a})}_{1} - {(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
If ${(E_{a})}_{1} > {(E_{a})}_{2}, t h e n T C_{1} < T C_{2}$ .

CHXII04:CHEMICAL KINETICS

320252 The activation energy for a simple chemical
reaction $A \to B$ is $E_{a}$ in forward direction. The
activation energy for reverse reaction

1 is always less than

E_{a}

2 can be less than or more than

E_{a}

3 is always double of

E_{a}

4 is negative of

E_{a}

Explanation:

Since the nature of reaction (i.e. exothermic or endothermic) is not given, $E_{a}$ for reverse reaction can be more or less.

CHXII04:CHEMICAL KINETICS

320253 The decomposition of $N_{2} O into N_{2}$ and in presence gaseous argon follow second order kinetics with $k = (5 .0 \times 1 0^{11} L m o l^{- 1} s^{- 1}) e^{- \frac{41570}{T}} K$ stands for Kelvin units). The energy of activation of the reaction is

1

5.0 \times 10^{11} J

2

41570 J

3

5000 J

4

345612.98 J

Explanation:

$E_{a} = 41570 \times R = 41570 \times 8 .31 = 345612 .98 J$

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320254 For a reaction, the value of rate constant at 300 $K$ is $6.0 \times 10^{5} s^{- 1}$ . The value of Arrhenius factor $A$ , at infinitely high temperature is:

1

\frac{6 \times 10^{- 5}}{300}

2

6 \times 10^{5}

3

6 \times 10^{5} \times e^{- Ea / 300 R}

4

e^{- Ea / 300 R}

Explanation:

$k = A e^{\frac{- Ea}{RT}}$
At $T = \infty; k = A$
$= 6 \times 10^{5} s^{- 1}$

KCET - 2023

CHXII04:CHEMICAL KINETICS

320255 A plot of $\ln k$ versus $\frac{1}{T}$ for a reaction gives the slope $- 2.0 \times 10^{3} K$ . The energy of activation for the reaction is ____ $kJ {mol}^{- 1}$ .

1 16.63

2 20.75

3 26.63

4 8.31

Explanation:

In a plot of $\ln k vs \frac{1}{T}$ ,
$-$ slope $= \frac{E_{a}}{R}$
$2 \times 10^{3} = \frac{E_{a}}{8.314}$
$E_{a} = 8.314 \times 2 \times 10^{3} = 16628 J = 16.63 kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320256 If for two reactions $E_{a_{1}} > E_{a_{2}}$ and ${TC}_{1}$ and ${TC}_{2}$ are temperature coefficient respectively, then which alternative is correct?

1

{TC}_{1} > {TC}_{2}

2

{TC}_{1} < {TC}_{2}

3

{TC}_{1} = {TC}_{2}

4 cannot be determined

Explanation:

We, know,
$l n (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$\frac{k_{2}}{k_{1}} = e^{\frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{1} = e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{2} = e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$\frac{T C_{1}}{T C_{2}} = \frac{e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}}{e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}} = e^{\frac{{(E_{a})}_{1} - {(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
If ${(E_{a})}_{1} > {(E_{a})}_{2}, t h e n T C_{1} < T C_{2}$ .

CHXII04:CHEMICAL KINETICS

320252 The activation energy for a simple chemical
reaction $A \to B$ is $E_{a}$ in forward direction. The
activation energy for reverse reaction

1 is always less than

E_{a}

2 can be less than or more than

E_{a}

3 is always double of

E_{a}

4 is negative of

E_{a}

Explanation:

Since the nature of reaction (i.e. exothermic or endothermic) is not given, $E_{a}$ for reverse reaction can be more or less.

CHXII04:CHEMICAL KINETICS

320253 The decomposition of $N_{2} O into N_{2}$ and in presence gaseous argon follow second order kinetics with $k = (5 .0 \times 1 0^{11} L m o l^{- 1} s^{- 1}) e^{- \frac{41570}{T}} K$ stands for Kelvin units). The energy of activation of the reaction is

1

5.0 \times 10^{11} J

2

41570 J

3

5000 J

4

345612.98 J

Explanation:

$E_{a} = 41570 \times R = 41570 \times 8 .31 = 345612 .98 J$

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320254 For a reaction, the value of rate constant at 300 $K$ is $6.0 \times 10^{5} s^{- 1}$ . The value of Arrhenius factor $A$ , at infinitely high temperature is:

1

\frac{6 \times 10^{- 5}}{300}

2

6 \times 10^{5}

3

6 \times 10^{5} \times e^{- Ea / 300 R}

4

e^{- Ea / 300 R}

Explanation:

$k = A e^{\frac{- Ea}{RT}}$
At $T = \infty; k = A$
$= 6 \times 10^{5} s^{- 1}$

KCET - 2023

CHXII04:CHEMICAL KINETICS

320255 A plot of $\ln k$ versus $\frac{1}{T}$ for a reaction gives the slope $- 2.0 \times 10^{3} K$ . The energy of activation for the reaction is ____ $kJ {mol}^{- 1}$ .

1 16.63

2 20.75

3 26.63

4 8.31

Explanation:

In a plot of $\ln k vs \frac{1}{T}$ ,
$-$ slope $= \frac{E_{a}}{R}$
$2 \times 10^{3} = \frac{E_{a}}{8.314}$
$E_{a} = 8.314 \times 2 \times 10^{3} = 16628 J = 16.63 kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320256 If for two reactions $E_{a_{1}} > E_{a_{2}}$ and ${TC}_{1}$ and ${TC}_{2}$ are temperature coefficient respectively, then which alternative is correct?

1

{TC}_{1} > {TC}_{2}

2

{TC}_{1} < {TC}_{2}

3

{TC}_{1} = {TC}_{2}

4 cannot be determined

Explanation:

We, know,
$l n (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$\frac{k_{2}}{k_{1}} = e^{\frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{1} = e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{2} = e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$\frac{T C_{1}}{T C_{2}} = \frac{e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}}{e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}} = e^{\frac{{(E_{a})}_{1} - {(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
If ${(E_{a})}_{1} > {(E_{a})}_{2}, t h e n T C_{1} < T C_{2}$ .

CHXII04:CHEMICAL KINETICS

320252 The activation energy for a simple chemical
reaction $A \to B$ is $E_{a}$ in forward direction. The
activation energy for reverse reaction

1 is always less than

E_{a}

2 can be less than or more than

E_{a}

3 is always double of

E_{a}

4 is negative of

E_{a}

Explanation:

Since the nature of reaction (i.e. exothermic or endothermic) is not given, $E_{a}$ for reverse reaction can be more or less.

CHXII04:CHEMICAL KINETICS

320253 The decomposition of $N_{2} O into N_{2}$ and in presence gaseous argon follow second order kinetics with $k = (5 .0 \times 1 0^{11} L m o l^{- 1} s^{- 1}) e^{- \frac{41570}{T}} K$ stands for Kelvin units). The energy of activation of the reaction is

1

5.0 \times 10^{11} J

2

41570 J

3

5000 J

4

345612.98 J

Explanation:

$E_{a} = 41570 \times R = 41570 \times 8 .31 = 345612 .98 J$

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320254 For a reaction, the value of rate constant at 300 $K$ is $6.0 \times 10^{5} s^{- 1}$ . The value of Arrhenius factor $A$ , at infinitely high temperature is:

1

\frac{6 \times 10^{- 5}}{300}

2

6 \times 10^{5}

3

6 \times 10^{5} \times e^{- Ea / 300 R}

4

e^{- Ea / 300 R}

Explanation:

$k = A e^{\frac{- Ea}{RT}}$
At $T = \infty; k = A$
$= 6 \times 10^{5} s^{- 1}$

KCET - 2023

CHXII04:CHEMICAL KINETICS

320255 A plot of $\ln k$ versus $\frac{1}{T}$ for a reaction gives the slope $- 2.0 \times 10^{3} K$ . The energy of activation for the reaction is ____ $kJ {mol}^{- 1}$ .

1 16.63

2 20.75

3 26.63

4 8.31

Explanation:

In a plot of $\ln k vs \frac{1}{T}$ ,
$-$ slope $= \frac{E_{a}}{R}$
$2 \times 10^{3} = \frac{E_{a}}{8.314}$
$E_{a} = 8.314 \times 2 \times 10^{3} = 16628 J = 16.63 kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320256 If for two reactions $E_{a_{1}} > E_{a_{2}}$ and ${TC}_{1}$ and ${TC}_{2}$ are temperature coefficient respectively, then which alternative is correct?

1

{TC}_{1} > {TC}_{2}

2

{TC}_{1} < {TC}_{2}

3

{TC}_{1} = {TC}_{2}

4 cannot be determined

Explanation:

We, know,
$l n (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$\frac{k_{2}}{k_{1}} = e^{\frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{1} = e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$T C_{2} = e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
$\frac{T C_{1}}{T C_{2}} = \frac{e^{\frac{{(E_{a})}_{1}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}}{e^{\frac{{(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}} = e^{\frac{{(E_{a})}_{1} - {(E_{a})}_{2}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})}$
If ${(E_{a})}_{1} > {(E_{a})}_{2}, t h e n T C_{1} < T C_{2}$ .