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CHXII04:CHEMICAL KINETICS

320506 The rates of a certain reaction $(dc / dt)$ at different times are as follows?
$\begin{matrix} T i m e & R a t e (m o l e l i t e r^{- 1} s e c^{- 1}) \\ 0 & 2 .8 \times 1 0^{- 2} \\ 10 & 2 .78 \times 1 0^{- 2} \\ 20 & 2 .81 \times 1 0^{- 2} \\ 30 & 2 .79 \times 1 0^{- 2} \end{matrix}$
The reaction is

1 Zero order

2 First order

3 Second order

4 Third order

CHXII04:CHEMICAL KINETICS

320507 For the reaction $aA \to xP$ when $[A] = 2 .2 mM$ the rate was found to be $2 .4 m M s^{- 1}$ . On reducing the concentration of A to half, the rate changes to $0 .6 mM s^{- 1}$ . The order of reaction with respect to $A$ is

1

1 .5

2

2 .0

3

2 .5

4

3 .0

Explanation:

$r = k[A]^{a} (1)$
$\frac{r}{4} = k {[\frac{A}{2}]}^{a} (2)$
$\frac{(2)}{(1)} \Rightarrow 4 = 2^{a}$
$a = 2$

CHXII04:CHEMICAL KINETICS

320508 A student has studied the decomposition of a gas ${AB}_{3}$ at $25^{\circ} C$ . He obtained the following data.
supporting img

The order of the reaction is

1 0

2 1

3 0.5

4 2

Explanation:

$t_{1 / 2} \propto {(C_{0})}^{1 - n}$
In terms of pressure,
$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{p_{1}}{p_{2}})}^{1 - n}$
$\frac{4}{2} = {(\frac{50}{100})}^{1 - n}$
or $(2)^{1} = {(\frac{1}{2})}^{1 - n}$
or $(2)^{1} = (2)^{n - 1}$
On equating, we get
$1 = n - 1$
$n = 2$

JEE - 2023

CHXII04:CHEMICAL KINETICS

320509 The following results were obtained during kinetic studies of the reaction; $2 A + B \to$ products
The time (in minutes) required to consume half of $A$ is
supporting img

1 5

2 1

3 10

4 100

Explanation:

Assume $r = k[A]^{x} {[B]}^{y}$
Taking exp. I and II, we get order of [B]
$\begin{aligned} {(\frac{0.20}{0.25})}^{y} = \frac{6.93 \times 10^{- 3}}{6.93 \times 10^{- 3}} = 1 \\ {(\frac{1}{2})}^{y} = {[\frac{1}{2}]}^{0} \Rightarrow y = 0 \end{aligned}$
Taking expanision I and III,
$\begin{aligned} \frac{6.93 \times 10^{- 3}}{1.386 \times 10^{- 2}} = {(\frac{1}{2})}^{x} \\ x = 1 \end{aligned}$
$∴$ Overall order $= 0 + 1 = 1$
Now, rate $= k [A]^{1} [B]^{0}$
$\begin{aligned} 6.93 \times 10^{- 3} = k [0.10] \\ k = \frac{6.93 \times 10^{- 3}}{0.10} = 0.0693 \min \\ t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0693} = 10 \min \end{aligned}$
To consume half of $A$ it becomes $\frac{10}{2} = 5 \min$

JEE - 2019

CHXII04:CHEMICAL KINETICS

320506 The rates of a certain reaction $(dc / dt)$ at different times are as follows?
$\begin{matrix} T i m e & R a t e (m o l e l i t e r^{- 1} s e c^{- 1}) \\ 0 & 2 .8 \times 1 0^{- 2} \\ 10 & 2 .78 \times 1 0^{- 2} \\ 20 & 2 .81 \times 1 0^{- 2} \\ 30 & 2 .79 \times 1 0^{- 2} \end{matrix}$
The reaction is

1 Zero order

2 First order

3 Second order

4 Third order

CHXII04:CHEMICAL KINETICS

320507 For the reaction $aA \to xP$ when $[A] = 2 .2 mM$ the rate was found to be $2 .4 m M s^{- 1}$ . On reducing the concentration of A to half, the rate changes to $0 .6 mM s^{- 1}$ . The order of reaction with respect to $A$ is

1

1 .5

2

2 .0

3

2 .5

4

3 .0

Explanation:

$r = k[A]^{a} (1)$
$\frac{r}{4} = k {[\frac{A}{2}]}^{a} (2)$
$\frac{(2)}{(1)} \Rightarrow 4 = 2^{a}$
$a = 2$

CHXII04:CHEMICAL KINETICS

320508 A student has studied the decomposition of a gas ${AB}_{3}$ at $25^{\circ} C$ . He obtained the following data.
supporting img

The order of the reaction is

1 0

2 1

3 0.5

4 2

Explanation:

$t_{1 / 2} \propto {(C_{0})}^{1 - n}$
In terms of pressure,
$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{p_{1}}{p_{2}})}^{1 - n}$
$\frac{4}{2} = {(\frac{50}{100})}^{1 - n}$
or $(2)^{1} = {(\frac{1}{2})}^{1 - n}$
or $(2)^{1} = (2)^{n - 1}$
On equating, we get
$1 = n - 1$
$n = 2$

JEE - 2023

CHXII04:CHEMICAL KINETICS

320509 The following results were obtained during kinetic studies of the reaction; $2 A + B \to$ products
The time (in minutes) required to consume half of $A$ is
supporting img

1 5

2 1

3 10

4 100

Explanation:

Assume $r = k[A]^{x} {[B]}^{y}$
Taking exp. I and II, we get order of [B]
$\begin{aligned} {(\frac{0.20}{0.25})}^{y} = \frac{6.93 \times 10^{- 3}}{6.93 \times 10^{- 3}} = 1 \\ {(\frac{1}{2})}^{y} = {[\frac{1}{2}]}^{0} \Rightarrow y = 0 \end{aligned}$
Taking expanision I and III,
$\begin{aligned} \frac{6.93 \times 10^{- 3}}{1.386 \times 10^{- 2}} = {(\frac{1}{2})}^{x} \\ x = 1 \end{aligned}$
$∴$ Overall order $= 0 + 1 = 1$
Now, rate $= k [A]^{1} [B]^{0}$
$\begin{aligned} 6.93 \times 10^{- 3} = k [0.10] \\ k = \frac{6.93 \times 10^{- 3}}{0.10} = 0.0693 \min \\ t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0693} = 10 \min \end{aligned}$
To consume half of $A$ it becomes $\frac{10}{2} = 5 \min$

JEE - 2019

CHXII04:CHEMICAL KINETICS

320506 The rates of a certain reaction $(dc / dt)$ at different times are as follows?
$\begin{matrix} T i m e & R a t e (m o l e l i t e r^{- 1} s e c^{- 1}) \\ 0 & 2 .8 \times 1 0^{- 2} \\ 10 & 2 .78 \times 1 0^{- 2} \\ 20 & 2 .81 \times 1 0^{- 2} \\ 30 & 2 .79 \times 1 0^{- 2} \end{matrix}$
The reaction is

1 Zero order

2 First order

3 Second order

4 Third order

CHXII04:CHEMICAL KINETICS

320507 For the reaction $aA \to xP$ when $[A] = 2 .2 mM$ the rate was found to be $2 .4 m M s^{- 1}$ . On reducing the concentration of A to half, the rate changes to $0 .6 mM s^{- 1}$ . The order of reaction with respect to $A$ is

1

1 .5

2

2 .0

3

2 .5

4

3 .0

Explanation:

$r = k[A]^{a} (1)$
$\frac{r}{4} = k {[\frac{A}{2}]}^{a} (2)$
$\frac{(2)}{(1)} \Rightarrow 4 = 2^{a}$
$a = 2$

CHXII04:CHEMICAL KINETICS

320508 A student has studied the decomposition of a gas ${AB}_{3}$ at $25^{\circ} C$ . He obtained the following data.
supporting img

The order of the reaction is

1 0

2 1

3 0.5

4 2

Explanation:

$t_{1 / 2} \propto {(C_{0})}^{1 - n}$
In terms of pressure,
$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{p_{1}}{p_{2}})}^{1 - n}$
$\frac{4}{2} = {(\frac{50}{100})}^{1 - n}$
or $(2)^{1} = {(\frac{1}{2})}^{1 - n}$
or $(2)^{1} = (2)^{n - 1}$
On equating, we get
$1 = n - 1$
$n = 2$

JEE - 2023

CHXII04:CHEMICAL KINETICS

320509 The following results were obtained during kinetic studies of the reaction; $2 A + B \to$ products
The time (in minutes) required to consume half of $A$ is
supporting img

1 5

2 1

3 10

4 100

Explanation:

Assume $r = k[A]^{x} {[B]}^{y}$
Taking exp. I and II, we get order of [B]
$\begin{aligned} {(\frac{0.20}{0.25})}^{y} = \frac{6.93 \times 10^{- 3}}{6.93 \times 10^{- 3}} = 1 \\ {(\frac{1}{2})}^{y} = {[\frac{1}{2}]}^{0} \Rightarrow y = 0 \end{aligned}$
Taking expanision I and III,
$\begin{aligned} \frac{6.93 \times 10^{- 3}}{1.386 \times 10^{- 2}} = {(\frac{1}{2})}^{x} \\ x = 1 \end{aligned}$
$∴$ Overall order $= 0 + 1 = 1$
Now, rate $= k [A]^{1} [B]^{0}$
$\begin{aligned} 6.93 \times 10^{- 3} = k [0.10] \\ k = \frac{6.93 \times 10^{- 3}}{0.10} = 0.0693 \min \\ t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0693} = 10 \min \end{aligned}$
To consume half of $A$ it becomes $\frac{10}{2} = 5 \min$

JEE - 2019

CHXII04:CHEMICAL KINETICS

320506 The rates of a certain reaction $(dc / dt)$ at different times are as follows?
$\begin{matrix} T i m e & R a t e (m o l e l i t e r^{- 1} s e c^{- 1}) \\ 0 & 2 .8 \times 1 0^{- 2} \\ 10 & 2 .78 \times 1 0^{- 2} \\ 20 & 2 .81 \times 1 0^{- 2} \\ 30 & 2 .79 \times 1 0^{- 2} \end{matrix}$
The reaction is

1 Zero order

2 First order

3 Second order

4 Third order

CHXII04:CHEMICAL KINETICS

320507 For the reaction $aA \to xP$ when $[A] = 2 .2 mM$ the rate was found to be $2 .4 m M s^{- 1}$ . On reducing the concentration of A to half, the rate changes to $0 .6 mM s^{- 1}$ . The order of reaction with respect to $A$ is

1

1 .5

2

2 .0

3

2 .5

4

3 .0

Explanation:

$r = k[A]^{a} (1)$
$\frac{r}{4} = k {[\frac{A}{2}]}^{a} (2)$
$\frac{(2)}{(1)} \Rightarrow 4 = 2^{a}$
$a = 2$

CHXII04:CHEMICAL KINETICS

320508 A student has studied the decomposition of a gas ${AB}_{3}$ at $25^{\circ} C$ . He obtained the following data.
supporting img

The order of the reaction is

1 0

2 1

3 0.5

4 2

Explanation:

$t_{1 / 2} \propto {(C_{0})}^{1 - n}$
In terms of pressure,
$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{p_{1}}{p_{2}})}^{1 - n}$
$\frac{4}{2} = {(\frac{50}{100})}^{1 - n}$
or $(2)^{1} = {(\frac{1}{2})}^{1 - n}$
or $(2)^{1} = (2)^{n - 1}$
On equating, we get
$1 = n - 1$
$n = 2$

JEE - 2023

CHXII04:CHEMICAL KINETICS

320509 The following results were obtained during kinetic studies of the reaction; $2 A + B \to$ products
The time (in minutes) required to consume half of $A$ is
supporting img

1 5

2 1

3 10

4 100

Explanation:

Assume $r = k[A]^{x} {[B]}^{y}$
Taking exp. I and II, we get order of [B]
$\begin{aligned} {(\frac{0.20}{0.25})}^{y} = \frac{6.93 \times 10^{- 3}}{6.93 \times 10^{- 3}} = 1 \\ {(\frac{1}{2})}^{y} = {[\frac{1}{2}]}^{0} \Rightarrow y = 0 \end{aligned}$
Taking expanision I and III,
$\begin{aligned} \frac{6.93 \times 10^{- 3}}{1.386 \times 10^{- 2}} = {(\frac{1}{2})}^{x} \\ x = 1 \end{aligned}$
$∴$ Overall order $= 0 + 1 = 1$
Now, rate $= k [A]^{1} [B]^{0}$
$\begin{aligned} 6.93 \times 10^{- 3} = k [0.10] \\ k = \frac{6.93 \times 10^{- 3}}{0.10} = 0.0693 \min \\ t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0693} = 10 \min \end{aligned}$
To consume half of $A$ it becomes $\frac{10}{2} = 5 \min$

JEE - 2019

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