QBManager

CHXII04:CHEMICAL KINETICS

320222 The slope of Arrhenius Plot $(l n k v s \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3} K$ . The value of $E_{a}$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314 {JK}^{- 1} {mol}^{- 1}$ ]

1

83.0 kJ {mol}^{- 1}

2

166 kJ {mol}^{- 1}

3

- 83 kJ {mol}^{- 1}

4

41.5 kJ {mol}^{- 1}

Explanation:

$\ln k = \frac{- E_{a}}{R T} + \ln A$
Slope $= \frac{- E_{a}}{R}$
$\begin{aligned} - 5 \times 10^{3} = \frac{- E_{a}}{R} \\ E_{a} = 5 \times 10^{3} \times R \\ = 5 \times 10^{3} \times 8.314 \\ = 41.57 \times 10^{3} J \\ = 41.57 {kJmol}^{- 1} \end{aligned}$

NEET - 2021

CHXII04:CHEMICAL KINETICS

320223 The rate constant for the first order decomposition of a certain reaction is described by the equation
$\ln k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4} K}{T}$ . The energy of activation for this reaction is

1

1.26 \times 1 0^{4} cal mo l^{-  1}

2

4 .29 \times 1 0^{4} cal mo l^{- 1}

3

3 .12 \times 1 0^{4} cal mo l^{- 1}

4

2 .50 \times 1 0^{4} cal mo l^{- 1}

Explanation:

Given that,
In $k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4}}{T} K (1)$
We know that,
In $k (s^{- 1}) = \ln A - \frac{E_{a}}{RT} (2)$
On comparing equation (1) and (2) we get
$\frac{E_{a}}{R} = 1.25 \times 10^{4} K$
$∴ E_{a} = 1.25 \times 10^{4} K \times R cal K^{- 1} mo l^{- 1}$
$= 1.25 \times 10^{4} K \times 2 cal K^{- 1} mo l^{- 1}$ $[∵ R = 2 cal K^{- 1} {mol}^{- 1}]$
$= 2.50 \times 10^{4} cal mo l^{- 1}$

AIIMS - 2018

CHXII04:CHEMICAL KINETICS

320224 An endothermic reaction $A \to B$ has $E_{a} v a l u e s a s 15 K . C a l / m o l$ and enthalpy change is $5 K . C a l / m o l e$ . The $E_{a} o f B \to A$ is ( in K.Cal/mole)

1 20

2 15

3 10

4 5

Explanation:

Given, $Δ H = 5 k c a l / m o l$
${(E_{a})}_{f o r w a r d} = 15 k c a l / m o l$
${(E_{a})}_{B a c k w a r d} = ?$
$∴ Δ H = {(E_{a})}_{f o r w a r d} - {(E_{a})}_{B a c k w a r d}$
$5 = 15 - {(E_{a})}_{B a c k w a r d}$
${(E_{a})}_{B a c k w a r d} = 10 k c a l m o l^{- 1}$

CHXII04:CHEMICAL KINETICS

320225 Slope of the straight line obtained by plotting $\log_{10} k$ against $1 / T$ represents what term?

1

- E_{a}

2

- 2 .303 E_{a} / R

3

- E_{a} / 2 .303 R

4

E_{a} / R

Explanation:

According to Arrhenius equation,
$k = {Ae}^{- E_{a} / RT}$
By taking log on both sides
$\log k = \log_{e} A - \frac{E_{a}}{RT}$
$\log k = \log A - \frac{1}{T} \cdot \frac{E_{a}}{2 \cdot 303 R}$
Comparing it with $y = mx + c$
A plot between $\log_{10} k$ and $(\frac{1}{T})$ is a straight line and slope is given by
supporting img
$Slope = \frac{- E_{a}}{2.303 R}$
(Slope has negative value)

KCET - 2019

CHXII04:CHEMICAL KINETICS

320226 Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is 1.75 . The activation energy for the saponification of ethyl acetate is

1

10 .7k cal mo l^{- 1}

2

12 .7k cal mo l^{- 1}

3

14 .7k cal mo l^{- 1}

4

16 .7k cal mo l^{- 1}

Explanation:

$m = \frac{k_{T + 10}}{k} = 1 .75$
$log \frac{k_{T + 10}}{k} = \frac{E_{a}}{2 .303R} [\frac{1}{25} - \frac{1}{35}]$
$l o g 1 .75 = \frac{E_{a}}{2 .303 \times 2} [\frac{10}{875}]$
$E_{a} = 10 .7k cal mo l^{- 1}$

CHXII04:CHEMICAL KINETICS

320222 The slope of Arrhenius Plot $(l n k v s \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3} K$ . The value of $E_{a}$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314 {JK}^{- 1} {mol}^{- 1}$ ]

1

83.0 kJ {mol}^{- 1}

2

166 kJ {mol}^{- 1}

3

- 83 kJ {mol}^{- 1}

4

41.5 kJ {mol}^{- 1}

Explanation:

$\ln k = \frac{- E_{a}}{R T} + \ln A$
Slope $= \frac{- E_{a}}{R}$
$\begin{aligned} - 5 \times 10^{3} = \frac{- E_{a}}{R} \\ E_{a} = 5 \times 10^{3} \times R \\ = 5 \times 10^{3} \times 8.314 \\ = 41.57 \times 10^{3} J \\ = 41.57 {kJmol}^{- 1} \end{aligned}$

NEET - 2021

CHXII04:CHEMICAL KINETICS

320223 The rate constant for the first order decomposition of a certain reaction is described by the equation
$\ln k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4} K}{T}$ . The energy of activation for this reaction is

1

1.26 \times 1 0^{4} cal mo l^{-  1}

2

4 .29 \times 1 0^{4} cal mo l^{- 1}

3

3 .12 \times 1 0^{4} cal mo l^{- 1}

4

2 .50 \times 1 0^{4} cal mo l^{- 1}

Explanation:

Given that,
In $k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4}}{T} K (1)$
We know that,
In $k (s^{- 1}) = \ln A - \frac{E_{a}}{RT} (2)$
On comparing equation (1) and (2) we get
$\frac{E_{a}}{R} = 1.25 \times 10^{4} K$
$∴ E_{a} = 1.25 \times 10^{4} K \times R cal K^{- 1} mo l^{- 1}$
$= 1.25 \times 10^{4} K \times 2 cal K^{- 1} mo l^{- 1}$ $[∵ R = 2 cal K^{- 1} {mol}^{- 1}]$
$= 2.50 \times 10^{4} cal mo l^{- 1}$

AIIMS - 2018

CHXII04:CHEMICAL KINETICS

320224 An endothermic reaction $A \to B$ has $E_{a} v a l u e s a s 15 K . C a l / m o l$ and enthalpy change is $5 K . C a l / m o l e$ . The $E_{a} o f B \to A$ is ( in K.Cal/mole)

1 20

2 15

3 10

4 5

Explanation:

Given, $Δ H = 5 k c a l / m o l$
${(E_{a})}_{f o r w a r d} = 15 k c a l / m o l$
${(E_{a})}_{B a c k w a r d} = ?$
$∴ Δ H = {(E_{a})}_{f o r w a r d} - {(E_{a})}_{B a c k w a r d}$
$5 = 15 - {(E_{a})}_{B a c k w a r d}$
${(E_{a})}_{B a c k w a r d} = 10 k c a l m o l^{- 1}$

CHXII04:CHEMICAL KINETICS

320225 Slope of the straight line obtained by plotting $\log_{10} k$ against $1 / T$ represents what term?

1

- E_{a}

2

- 2 .303 E_{a} / R

3

- E_{a} / 2 .303 R

4

E_{a} / R

Explanation:

According to Arrhenius equation,
$k = {Ae}^{- E_{a} / RT}$
By taking log on both sides
$\log k = \log_{e} A - \frac{E_{a}}{RT}$
$\log k = \log A - \frac{1}{T} \cdot \frac{E_{a}}{2 \cdot 303 R}$
Comparing it with $y = mx + c$
A plot between $\log_{10} k$ and $(\frac{1}{T})$ is a straight line and slope is given by
supporting img
$Slope = \frac{- E_{a}}{2.303 R}$
(Slope has negative value)

KCET - 2019

CHXII04:CHEMICAL KINETICS

320226 Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is 1.75 . The activation energy for the saponification of ethyl acetate is

1

10 .7k cal mo l^{- 1}

2

12 .7k cal mo l^{- 1}

3

14 .7k cal mo l^{- 1}

4

16 .7k cal mo l^{- 1}

Explanation:

$m = \frac{k_{T + 10}}{k} = 1 .75$
$log \frac{k_{T + 10}}{k} = \frac{E_{a}}{2 .303R} [\frac{1}{25} - \frac{1}{35}]$
$l o g 1 .75 = \frac{E_{a}}{2 .303 \times 2} [\frac{10}{875}]$
$E_{a} = 10 .7k cal mo l^{- 1}$

CHXII04:CHEMICAL KINETICS

320222 The slope of Arrhenius Plot $(l n k v s \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3} K$ . The value of $E_{a}$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314 {JK}^{- 1} {mol}^{- 1}$ ]

1

83.0 kJ {mol}^{- 1}

2

166 kJ {mol}^{- 1}

3

- 83 kJ {mol}^{- 1}

4

41.5 kJ {mol}^{- 1}

Explanation:

$\ln k = \frac{- E_{a}}{R T} + \ln A$
Slope $= \frac{- E_{a}}{R}$
$\begin{aligned} - 5 \times 10^{3} = \frac{- E_{a}}{R} \\ E_{a} = 5 \times 10^{3} \times R \\ = 5 \times 10^{3} \times 8.314 \\ = 41.57 \times 10^{3} J \\ = 41.57 {kJmol}^{- 1} \end{aligned}$

NEET - 2021

CHXII04:CHEMICAL KINETICS

320223 The rate constant for the first order decomposition of a certain reaction is described by the equation
$\ln k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4} K}{T}$ . The energy of activation for this reaction is

1

1.26 \times 1 0^{4} cal mo l^{-  1}

2

4 .29 \times 1 0^{4} cal mo l^{- 1}

3

3 .12 \times 1 0^{4} cal mo l^{- 1}

4

2 .50 \times 1 0^{4} cal mo l^{- 1}

Explanation:

Given that,
In $k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4}}{T} K (1)$
We know that,
In $k (s^{- 1}) = \ln A - \frac{E_{a}}{RT} (2)$
On comparing equation (1) and (2) we get
$\frac{E_{a}}{R} = 1.25 \times 10^{4} K$
$∴ E_{a} = 1.25 \times 10^{4} K \times R cal K^{- 1} mo l^{- 1}$
$= 1.25 \times 10^{4} K \times 2 cal K^{- 1} mo l^{- 1}$ $[∵ R = 2 cal K^{- 1} {mol}^{- 1}]$
$= 2.50 \times 10^{4} cal mo l^{- 1}$

AIIMS - 2018

CHXII04:CHEMICAL KINETICS

320224 An endothermic reaction $A \to B$ has $E_{a} v a l u e s a s 15 K . C a l / m o l$ and enthalpy change is $5 K . C a l / m o l e$ . The $E_{a} o f B \to A$ is ( in K.Cal/mole)

1 20

2 15

3 10

4 5

Explanation:

Given, $Δ H = 5 k c a l / m o l$
${(E_{a})}_{f o r w a r d} = 15 k c a l / m o l$
${(E_{a})}_{B a c k w a r d} = ?$
$∴ Δ H = {(E_{a})}_{f o r w a r d} - {(E_{a})}_{B a c k w a r d}$
$5 = 15 - {(E_{a})}_{B a c k w a r d}$
${(E_{a})}_{B a c k w a r d} = 10 k c a l m o l^{- 1}$

CHXII04:CHEMICAL KINETICS

320225 Slope of the straight line obtained by plotting $\log_{10} k$ against $1 / T$ represents what term?

1

- E_{a}

2

- 2 .303 E_{a} / R

3

- E_{a} / 2 .303 R

4

E_{a} / R

Explanation:

According to Arrhenius equation,
$k = {Ae}^{- E_{a} / RT}$
By taking log on both sides
$\log k = \log_{e} A - \frac{E_{a}}{RT}$
$\log k = \log A - \frac{1}{T} \cdot \frac{E_{a}}{2 \cdot 303 R}$
Comparing it with $y = mx + c$
A plot between $\log_{10} k$ and $(\frac{1}{T})$ is a straight line and slope is given by
supporting img
$Slope = \frac{- E_{a}}{2.303 R}$
(Slope has negative value)

KCET - 2019

CHXII04:CHEMICAL KINETICS

320226 Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is 1.75 . The activation energy for the saponification of ethyl acetate is

1

10 .7k cal mo l^{- 1}

2

12 .7k cal mo l^{- 1}

3

14 .7k cal mo l^{- 1}

4

16 .7k cal mo l^{- 1}

Explanation:

$m = \frac{k_{T + 10}}{k} = 1 .75$
$log \frac{k_{T + 10}}{k} = \frac{E_{a}}{2 .303R} [\frac{1}{25} - \frac{1}{35}]$
$l o g 1 .75 = \frac{E_{a}}{2 .303 \times 2} [\frac{10}{875}]$
$E_{a} = 10 .7k cal mo l^{- 1}$

CHXII04:CHEMICAL KINETICS

320222 The slope of Arrhenius Plot $(l n k v s \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3} K$ . The value of $E_{a}$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314 {JK}^{- 1} {mol}^{- 1}$ ]

1

83.0 kJ {mol}^{- 1}

2

166 kJ {mol}^{- 1}

3

- 83 kJ {mol}^{- 1}

4

41.5 kJ {mol}^{- 1}

Explanation:

$\ln k = \frac{- E_{a}}{R T} + \ln A$
Slope $= \frac{- E_{a}}{R}$
$\begin{aligned} - 5 \times 10^{3} = \frac{- E_{a}}{R} \\ E_{a} = 5 \times 10^{3} \times R \\ = 5 \times 10^{3} \times 8.314 \\ = 41.57 \times 10^{3} J \\ = 41.57 {kJmol}^{- 1} \end{aligned}$

NEET - 2021

CHXII04:CHEMICAL KINETICS

320223 The rate constant for the first order decomposition of a certain reaction is described by the equation
$\ln k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4} K}{T}$ . The energy of activation for this reaction is

1

1.26 \times 1 0^{4} cal mo l^{-  1}

2

4 .29 \times 1 0^{4} cal mo l^{- 1}

3

3 .12 \times 1 0^{4} cal mo l^{- 1}

4

2 .50 \times 1 0^{4} cal mo l^{- 1}

Explanation:

Given that,
In $k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4}}{T} K (1)$
We know that,
In $k (s^{- 1}) = \ln A - \frac{E_{a}}{RT} (2)$
On comparing equation (1) and (2) we get
$\frac{E_{a}}{R} = 1.25 \times 10^{4} K$
$∴ E_{a} = 1.25 \times 10^{4} K \times R cal K^{- 1} mo l^{- 1}$
$= 1.25 \times 10^{4} K \times 2 cal K^{- 1} mo l^{- 1}$ $[∵ R = 2 cal K^{- 1} {mol}^{- 1}]$
$= 2.50 \times 10^{4} cal mo l^{- 1}$

AIIMS - 2018

CHXII04:CHEMICAL KINETICS

320224 An endothermic reaction $A \to B$ has $E_{a} v a l u e s a s 15 K . C a l / m o l$ and enthalpy change is $5 K . C a l / m o l e$ . The $E_{a} o f B \to A$ is ( in K.Cal/mole)

1 20

2 15

3 10

4 5

Explanation:

Given, $Δ H = 5 k c a l / m o l$
${(E_{a})}_{f o r w a r d} = 15 k c a l / m o l$
${(E_{a})}_{B a c k w a r d} = ?$
$∴ Δ H = {(E_{a})}_{f o r w a r d} - {(E_{a})}_{B a c k w a r d}$
$5 = 15 - {(E_{a})}_{B a c k w a r d}$
${(E_{a})}_{B a c k w a r d} = 10 k c a l m o l^{- 1}$

CHXII04:CHEMICAL KINETICS

320225 Slope of the straight line obtained by plotting $\log_{10} k$ against $1 / T$ represents what term?

1

- E_{a}

2

- 2 .303 E_{a} / R

3

- E_{a} / 2 .303 R

4

E_{a} / R

Explanation:

According to Arrhenius equation,
$k = {Ae}^{- E_{a} / RT}$
By taking log on both sides
$\log k = \log_{e} A - \frac{E_{a}}{RT}$
$\log k = \log A - \frac{1}{T} \cdot \frac{E_{a}}{2 \cdot 303 R}$
Comparing it with $y = mx + c$
A plot between $\log_{10} k$ and $(\frac{1}{T})$ is a straight line and slope is given by
supporting img
$Slope = \frac{- E_{a}}{2.303 R}$
(Slope has negative value)

KCET - 2019

CHXII04:CHEMICAL KINETICS

320226 Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is 1.75 . The activation energy for the saponification of ethyl acetate is

1

10 .7k cal mo l^{- 1}

2

12 .7k cal mo l^{- 1}

3

14 .7k cal mo l^{- 1}

4

16 .7k cal mo l^{- 1}

Explanation:

$m = \frac{k_{T + 10}}{k} = 1 .75$
$log \frac{k_{T + 10}}{k} = \frac{E_{a}}{2 .303R} [\frac{1}{25} - \frac{1}{35}]$
$l o g 1 .75 = \frac{E_{a}}{2 .303 \times 2} [\frac{10}{875}]$
$E_{a} = 10 .7k cal mo l^{- 1}$

CHXII04:CHEMICAL KINETICS

320222 The slope of Arrhenius Plot $(l n k v s \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3} K$ . The value of $E_{a}$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314 {JK}^{- 1} {mol}^{- 1}$ ]

1

83.0 kJ {mol}^{- 1}

2

166 kJ {mol}^{- 1}

3

- 83 kJ {mol}^{- 1}

4

41.5 kJ {mol}^{- 1}

Explanation:

$\ln k = \frac{- E_{a}}{R T} + \ln A$
Slope $= \frac{- E_{a}}{R}$
$\begin{aligned} - 5 \times 10^{3} = \frac{- E_{a}}{R} \\ E_{a} = 5 \times 10^{3} \times R \\ = 5 \times 10^{3} \times 8.314 \\ = 41.57 \times 10^{3} J \\ = 41.57 {kJmol}^{- 1} \end{aligned}$

NEET - 2021

CHXII04:CHEMICAL KINETICS

320223 The rate constant for the first order decomposition of a certain reaction is described by the equation
$\ln k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4} K}{T}$ . The energy of activation for this reaction is

1

1.26 \times 1 0^{4} cal mo l^{-  1}

2

4 .29 \times 1 0^{4} cal mo l^{- 1}

3

3 .12 \times 1 0^{4} cal mo l^{- 1}

4

2 .50 \times 1 0^{4} cal mo l^{- 1}

Explanation:

Given that,
In $k (s^{- 1}) = 14.34 - \frac{1.25 \times 10^{4}}{T} K (1)$
We know that,
In $k (s^{- 1}) = \ln A - \frac{E_{a}}{RT} (2)$
On comparing equation (1) and (2) we get
$\frac{E_{a}}{R} = 1.25 \times 10^{4} K$
$∴ E_{a} = 1.25 \times 10^{4} K \times R cal K^{- 1} mo l^{- 1}$
$= 1.25 \times 10^{4} K \times 2 cal K^{- 1} mo l^{- 1}$ $[∵ R = 2 cal K^{- 1} {mol}^{- 1}]$
$= 2.50 \times 10^{4} cal mo l^{- 1}$

AIIMS - 2018

CHXII04:CHEMICAL KINETICS

320224 An endothermic reaction $A \to B$ has $E_{a} v a l u e s a s 15 K . C a l / m o l$ and enthalpy change is $5 K . C a l / m o l e$ . The $E_{a} o f B \to A$ is ( in K.Cal/mole)

1 20

2 15

3 10

4 5

Explanation:

Given, $Δ H = 5 k c a l / m o l$
${(E_{a})}_{f o r w a r d} = 15 k c a l / m o l$
${(E_{a})}_{B a c k w a r d} = ?$
$∴ Δ H = {(E_{a})}_{f o r w a r d} - {(E_{a})}_{B a c k w a r d}$
$5 = 15 - {(E_{a})}_{B a c k w a r d}$
${(E_{a})}_{B a c k w a r d} = 10 k c a l m o l^{- 1}$

CHXII04:CHEMICAL KINETICS

320225 Slope of the straight line obtained by plotting $\log_{10} k$ against $1 / T$ represents what term?

1

- E_{a}

2

- 2 .303 E_{a} / R

3

- E_{a} / 2 .303 R

4

E_{a} / R

Explanation:

According to Arrhenius equation,
$k = {Ae}^{- E_{a} / RT}$
By taking log on both sides
$\log k = \log_{e} A - \frac{E_{a}}{RT}$
$\log k = \log A - \frac{1}{T} \cdot \frac{E_{a}}{2 \cdot 303 R}$
Comparing it with $y = mx + c$
A plot between $\log_{10} k$ and $(\frac{1}{T})$ is a straight line and slope is given by
supporting img
$Slope = \frac{- E_{a}}{2.303 R}$
(Slope has negative value)

KCET - 2019

CHXII04:CHEMICAL KINETICS

320226 Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is 1.75 . The activation energy for the saponification of ethyl acetate is

1

10 .7k cal mo l^{- 1}

2

12 .7k cal mo l^{- 1}

3

14 .7k cal mo l^{- 1}

4

16 .7k cal mo l^{- 1}

Explanation:

$m = \frac{k_{T + 10}}{k} = 1 .75$
$log \frac{k_{T + 10}}{k} = \frac{E_{a}}{2 .303R} [\frac{1}{25} - \frac{1}{35}]$
$l o g 1 .75 = \frac{E_{a}}{2 .303 \times 2} [\frac{10}{875}]$
$E_{a} = 10 .7k cal mo l^{- 1}$

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