QBManager

Chemical Equilibrium

228993 The equilibrium constants for the reactions
$N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$
and $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {N H}_{3}$
are $K_{1}$ and $K_{2}$ respectively. Which one of the following is the correct relationship?

1

K_{1} = 2 K_{2}

2

K_{1} = \frac{1}{2} K_{2}

3

K_{2} = \sqrt{K_{1}}

4

K_{1} = K_{2}

Explanation:

The reaction is -
$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$
$K_{1} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} \dots \dots \dots \dots \dots \dots$
$\begin{aligned} \frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {NH}_{3} \\ K_{2} = \frac{[{NH}_{3}]}{{[N_{2}]}^{1 / 2} {[H_{2}]}^{3 / 2}} \end{aligned}$
Squaring on both sides -
$K_{2}^{2} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
From (i) \& (ii) we get
$K_{2}^{2} = K_{1} \Rightarrow K_{2} = \sqrt{K_{1}}$

BCECE-2011

Chemical Equilibrium

228996 In the reversible reaction.
$2 {NO}_{2} \underset{K_{2}}{\overset{K_{1}}{⇌}} N_{2} O_{4}$
the rate of disappearance of ${NO}_{2}$ is equal to

1

\frac{2 k_{1}}{k_{2}} {[{NO}_{2}]}^{2}

2

2 k_{1} {[{NO}_{2}]}^{2} - 2 k_{2} [N_{2} O_{4}]

3

2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]

4

(2 k_{1} - k_{2}) [{NO}_{2}]

Explanation:

In the reversible reaction
$2 {NO}_{2} ⇌ N_{2} O_{4}$
The concentration of ${NO}_{2}$ is decrease in the forward reaction and increase in backward reaction.
The rate of disappearance of ${NO}_{2}$ in the forward reaction is $- \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = k_{1} {[{NO}_{2}]}^{2}$
$- \frac{d [{NO}_{2}]}{dt} = 2 k_{1} {[{NO}_{2}]}^{2}$
Rate of formation of ${NO}_{2}$ in the reversible reaction is $\frac{d [{NO}_{2}]}{dt} = k_{2} [N_{2} O_{4}]$
$∴$ rate of disappearance of
${NO}_{2} = 2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]$

VITEEE- 2011

Chemical Equilibrium

228997 The unit of equilibrium constant for the following reaction is
$2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g)$

1

mol L^{- 1}

2

L {mol}^{- 1}

3

{mol}^{- 1} L^{- 1}

4 equilibrium constant is unitless

Explanation:

$\begin{aligned} 2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g) \\ K_{c} = \frac{[N_{2} O_{4}]}{{[{NO}_{2}]}^{2}} = \frac{{molL}^{- 1}}{{({molL}^{- 1})}^{2}} = L {mol}^{- 1} \end{aligned}$

UPTU/UPSEE-2011

Chemical Equilibrium

228998 If the equilibrium constant for reaction (i) is 2 , what is the equilibrium constant for reaction (ii)?
$\begin{aligned} {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \\ 2 C O + O_{2} ⇌ 2 {CO}_{2} \end{aligned}$

1

\frac{1}{4}

2

\frac{1}{2}

3 1

4 2

Explanation:

${CO}_{2} ⇌ CO + \frac{1}{2} O_{2}$
The equilibrium constant for the reaction-
$\begin{aligned} K_{c} = \frac{[CO] {[O_{2}]}^{1 / 2}}{[{CO}_{2}]} \\ {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \end{aligned}$
The equilibrium constant for the reaction-
$K_{c} = \frac{{[{CO}_{2}]}^{2}}{[CO]^{2} [O_{2}]} = {(\frac{1}{K_{c}})}^{2} = \frac{1}{2^{2}} = \frac{1}{4}$

UPTU/UPSEE-2011

Chemical Equilibrium

228993 The equilibrium constants for the reactions
$N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$
and $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {N H}_{3}$
are $K_{1}$ and $K_{2}$ respectively. Which one of the following is the correct relationship?

1

K_{1} = 2 K_{2}

2

K_{1} = \frac{1}{2} K_{2}

3

K_{2} = \sqrt{K_{1}}

4

K_{1} = K_{2}

Explanation:

The reaction is -
$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$
$K_{1} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} \dots \dots \dots \dots \dots \dots$
$\begin{aligned} \frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {NH}_{3} \\ K_{2} = \frac{[{NH}_{3}]}{{[N_{2}]}^{1 / 2} {[H_{2}]}^{3 / 2}} \end{aligned}$
Squaring on both sides -
$K_{2}^{2} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
From (i) \& (ii) we get
$K_{2}^{2} = K_{1} \Rightarrow K_{2} = \sqrt{K_{1}}$

BCECE-2011

Chemical Equilibrium

228996 In the reversible reaction.
$2 {NO}_{2} \underset{K_{2}}{\overset{K_{1}}{⇌}} N_{2} O_{4}$
the rate of disappearance of ${NO}_{2}$ is equal to

1

\frac{2 k_{1}}{k_{2}} {[{NO}_{2}]}^{2}

2

2 k_{1} {[{NO}_{2}]}^{2} - 2 k_{2} [N_{2} O_{4}]

3

2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]

4

(2 k_{1} - k_{2}) [{NO}_{2}]

Explanation:

In the reversible reaction
$2 {NO}_{2} ⇌ N_{2} O_{4}$
The concentration of ${NO}_{2}$ is decrease in the forward reaction and increase in backward reaction.
The rate of disappearance of ${NO}_{2}$ in the forward reaction is $- \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = k_{1} {[{NO}_{2}]}^{2}$
$- \frac{d [{NO}_{2}]}{dt} = 2 k_{1} {[{NO}_{2}]}^{2}$
Rate of formation of ${NO}_{2}$ in the reversible reaction is $\frac{d [{NO}_{2}]}{dt} = k_{2} [N_{2} O_{4}]$
$∴$ rate of disappearance of
${NO}_{2} = 2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]$

VITEEE- 2011

Chemical Equilibrium

228997 The unit of equilibrium constant for the following reaction is
$2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g)$

1

mol L^{- 1}

2

L {mol}^{- 1}

3

{mol}^{- 1} L^{- 1}

4 equilibrium constant is unitless

Explanation:

$\begin{aligned} 2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g) \\ K_{c} = \frac{[N_{2} O_{4}]}{{[{NO}_{2}]}^{2}} = \frac{{molL}^{- 1}}{{({molL}^{- 1})}^{2}} = L {mol}^{- 1} \end{aligned}$

UPTU/UPSEE-2011

Chemical Equilibrium

228998 If the equilibrium constant for reaction (i) is 2 , what is the equilibrium constant for reaction (ii)?
$\begin{aligned} {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \\ 2 C O + O_{2} ⇌ 2 {CO}_{2} \end{aligned}$

1

\frac{1}{4}

2

\frac{1}{2}

3 1

4 2

Explanation:

${CO}_{2} ⇌ CO + \frac{1}{2} O_{2}$
The equilibrium constant for the reaction-
$\begin{aligned} K_{c} = \frac{[CO] {[O_{2}]}^{1 / 2}}{[{CO}_{2}]} \\ {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \end{aligned}$
The equilibrium constant for the reaction-
$K_{c} = \frac{{[{CO}_{2}]}^{2}}{[CO]^{2} [O_{2}]} = {(\frac{1}{K_{c}})}^{2} = \frac{1}{2^{2}} = \frac{1}{4}$

UPTU/UPSEE-2011

Chemical Equilibrium

228993 The equilibrium constants for the reactions
$N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$
and $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {N H}_{3}$
are $K_{1}$ and $K_{2}$ respectively. Which one of the following is the correct relationship?

1

K_{1} = 2 K_{2}

2

K_{1} = \frac{1}{2} K_{2}

3

K_{2} = \sqrt{K_{1}}

4

K_{1} = K_{2}

Explanation:

The reaction is -
$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$
$K_{1} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} \dots \dots \dots \dots \dots \dots$
$\begin{aligned} \frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {NH}_{3} \\ K_{2} = \frac{[{NH}_{3}]}{{[N_{2}]}^{1 / 2} {[H_{2}]}^{3 / 2}} \end{aligned}$
Squaring on both sides -
$K_{2}^{2} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
From (i) \& (ii) we get
$K_{2}^{2} = K_{1} \Rightarrow K_{2} = \sqrt{K_{1}}$

BCECE-2011

Chemical Equilibrium

228996 In the reversible reaction.
$2 {NO}_{2} \underset{K_{2}}{\overset{K_{1}}{⇌}} N_{2} O_{4}$
the rate of disappearance of ${NO}_{2}$ is equal to

1

\frac{2 k_{1}}{k_{2}} {[{NO}_{2}]}^{2}

2

2 k_{1} {[{NO}_{2}]}^{2} - 2 k_{2} [N_{2} O_{4}]

3

2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]

4

(2 k_{1} - k_{2}) [{NO}_{2}]

Explanation:

In the reversible reaction
$2 {NO}_{2} ⇌ N_{2} O_{4}$
The concentration of ${NO}_{2}$ is decrease in the forward reaction and increase in backward reaction.
The rate of disappearance of ${NO}_{2}$ in the forward reaction is $- \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = k_{1} {[{NO}_{2}]}^{2}$
$- \frac{d [{NO}_{2}]}{dt} = 2 k_{1} {[{NO}_{2}]}^{2}$
Rate of formation of ${NO}_{2}$ in the reversible reaction is $\frac{d [{NO}_{2}]}{dt} = k_{2} [N_{2} O_{4}]$
$∴$ rate of disappearance of
${NO}_{2} = 2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]$

VITEEE- 2011

Chemical Equilibrium

228997 The unit of equilibrium constant for the following reaction is
$2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g)$

1

mol L^{- 1}

2

L {mol}^{- 1}

3

{mol}^{- 1} L^{- 1}

4 equilibrium constant is unitless

Explanation:

$\begin{aligned} 2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g) \\ K_{c} = \frac{[N_{2} O_{4}]}{{[{NO}_{2}]}^{2}} = \frac{{molL}^{- 1}}{{({molL}^{- 1})}^{2}} = L {mol}^{- 1} \end{aligned}$

UPTU/UPSEE-2011

Chemical Equilibrium

228998 If the equilibrium constant for reaction (i) is 2 , what is the equilibrium constant for reaction (ii)?
$\begin{aligned} {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \\ 2 C O + O_{2} ⇌ 2 {CO}_{2} \end{aligned}$

1

\frac{1}{4}

2

\frac{1}{2}

3 1

4 2

Explanation:

${CO}_{2} ⇌ CO + \frac{1}{2} O_{2}$
The equilibrium constant for the reaction-
$\begin{aligned} K_{c} = \frac{[CO] {[O_{2}]}^{1 / 2}}{[{CO}_{2}]} \\ {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \end{aligned}$
The equilibrium constant for the reaction-
$K_{c} = \frac{{[{CO}_{2}]}^{2}}{[CO]^{2} [O_{2}]} = {(\frac{1}{K_{c}})}^{2} = \frac{1}{2^{2}} = \frac{1}{4}$

UPTU/UPSEE-2011

Chemical Equilibrium

228993 The equilibrium constants for the reactions
$N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$
and $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {N H}_{3}$
are $K_{1}$ and $K_{2}$ respectively. Which one of the following is the correct relationship?

1

K_{1} = 2 K_{2}

2

K_{1} = \frac{1}{2} K_{2}

3

K_{2} = \sqrt{K_{1}}

4

K_{1} = K_{2}

Explanation:

The reaction is -
$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$
$K_{1} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} \dots \dots \dots \dots \dots \dots$
$\begin{aligned} \frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ {NH}_{3} \\ K_{2} = \frac{[{NH}_{3}]}{{[N_{2}]}^{1 / 2} {[H_{2}]}^{3 / 2}} \end{aligned}$
Squaring on both sides -
$K_{2}^{2} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
From (i) \& (ii) we get
$K_{2}^{2} = K_{1} \Rightarrow K_{2} = \sqrt{K_{1}}$

BCECE-2011

Chemical Equilibrium

228996 In the reversible reaction.
$2 {NO}_{2} \underset{K_{2}}{\overset{K_{1}}{⇌}} N_{2} O_{4}$
the rate of disappearance of ${NO}_{2}$ is equal to

1

\frac{2 k_{1}}{k_{2}} {[{NO}_{2}]}^{2}

2

2 k_{1} {[{NO}_{2}]}^{2} - 2 k_{2} [N_{2} O_{4}]

3

2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]

4

(2 k_{1} - k_{2}) [{NO}_{2}]

Explanation:

In the reversible reaction
$2 {NO}_{2} ⇌ N_{2} O_{4}$
The concentration of ${NO}_{2}$ is decrease in the forward reaction and increase in backward reaction.
The rate of disappearance of ${NO}_{2}$ in the forward reaction is $- \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = k_{1} {[{NO}_{2}]}^{2}$
$- \frac{d [{NO}_{2}]}{dt} = 2 k_{1} {[{NO}_{2}]}^{2}$
Rate of formation of ${NO}_{2}$ in the reversible reaction is $\frac{d [{NO}_{2}]}{dt} = k_{2} [N_{2} O_{4}]$
$∴$ rate of disappearance of
${NO}_{2} = 2 k_{1} {[{NO}_{2}]}^{2} - k_{2} [N_{2} O_{4}]$

VITEEE- 2011

Chemical Equilibrium

228997 The unit of equilibrium constant for the following reaction is
$2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g)$

1

mol L^{- 1}

2

L {mol}^{- 1}

3

{mol}^{- 1} L^{- 1}

4 equilibrium constant is unitless

Explanation:

$\begin{aligned} 2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g) \\ K_{c} = \frac{[N_{2} O_{4}]}{{[{NO}_{2}]}^{2}} = \frac{{molL}^{- 1}}{{({molL}^{- 1})}^{2}} = L {mol}^{- 1} \end{aligned}$

UPTU/UPSEE-2011

Chemical Equilibrium

228998 If the equilibrium constant for reaction (i) is 2 , what is the equilibrium constant for reaction (ii)?
$\begin{aligned} {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \\ 2 C O + O_{2} ⇌ 2 {CO}_{2} \end{aligned}$

1

\frac{1}{4}

2

\frac{1}{2}

3 1

4 2

Explanation:

${CO}_{2} ⇌ CO + \frac{1}{2} O_{2}$
The equilibrium constant for the reaction-
$\begin{aligned} K_{c} = \frac{[CO] {[O_{2}]}^{1 / 2}}{[{CO}_{2}]} \\ {CO}_{2} ⇌ CO + \frac{1}{2} O_{2} \end{aligned}$
The equilibrium constant for the reaction-
$K_{c} = \frac{{[{CO}_{2}]}^{2}}{[CO]^{2} [O_{2}]} = {(\frac{1}{K_{c}})}^{2} = \frac{1}{2^{2}} = \frac{1}{4}$

UPTU/UPSEE-2011