QBManager

Chemical Equilibrium

229234 Identify the correct expression for the equilibrium constant of the following reaction:
$2 X (g) + Y (g) ⇌ 3 Z (g)$

1

K = \frac{[X]^{2} [Y]}{[Z]^{3}}

2

K = \frac{[Z]^{3}}{[X]^{2} [Y]}

3

K = \frac{3 [Z]}{2 [X] [Y]}

4

K = [Z]^{3} [X]^{2} [Y]

Explanation:

Equilibrium constant is defined as, the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction.
Given that, $2 X (g) + Y (g)$ \rightleftharpoons 3Z $(g)$
Equilibrium constant $(K_{c}) = \frac{[Product]}{[Reactant]}$
$K_{c} = \frac{[Z]^{3}}{[X]^{2} [Y]}$

Shift-I

Chemical Equilibrium

229244 For the reaction taking place at certain temperature
${NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g)$ ,
if equilibrium pressure is $3 X$ bar then $Δ_{r} G^{0}$ would be

1

- RT \ln 9 - 3 RT \ln X

2

RT \ln 4 - 3 RT \ln X

3

- 3 RT \ln 4 X

4 None of these

Explanation:

From the relation b/w standard Gibb's free energy and equilibrium constant-
$Δ G^{o} = - RT \ln K_{p}; K_{p} = (2 X)^{2} X = 4 X^{3}$
$∴ Δ G^{o} = - RT \ln (4 X^{3})$
$Δ G^{o} = - RT \ln 4 - 3 RT \ln X$

BITSAT 2018

Chemical Equilibrium

229252 The value of $K_{c}$ for the dissociation reaction
$H_{2} (g) ⇌ 2 H (g) is 2.1 \times 10^{- 42}$
This equilibrium mixture contains mainly

1

H_{2} (g)

2

H (g)

atom

3

1 : 1

molar mixture of

H_{2}

and

H

4 1:2 molar mixture of

H_{2}

and

H

Explanation:

For the dissociation reaction -
$H_{2} (g) ⇌ 2 H (g) K_{c} = 2.1 \times 10^{- 42}$
Since, $K_{c}$ is very small.
Hence, $[H_{2}]$ is predominant.

UPTU/UPSEE-2015

Chemical Equilibrium

229255 If $K_{c}$ is the equilibrium constant for the formation of ${NH}_{3}$ , the dissociation constant of ${NH}_{3}$ under the same condition will be

1

\frac{1}{K_{c}}

2

K_{c}^{2}

3

\sqrt{K_{c}}

4

K_{c}

Explanation:

The equilibrium reaction is-
$N_{2} + 3 H_{2} \underset{k_{c}}{\overset{K_{c}}{⇌}} 2 {NH}_{3}$
If the equation representing equilibrium constant is reversed. The new equilibrium constant is inverse of the first value.
$i.e. K_{c}^{'} = \frac{1}{K_{c}}$

CG PET- 2015

Chemical Equilibrium

229234 Identify the correct expression for the equilibrium constant of the following reaction:
$2 X (g) + Y (g) ⇌ 3 Z (g)$

1

K = \frac{[X]^{2} [Y]}{[Z]^{3}}

2

K = \frac{[Z]^{3}}{[X]^{2} [Y]}

3

K = \frac{3 [Z]}{2 [X] [Y]}

4

K = [Z]^{3} [X]^{2} [Y]

Explanation:

Equilibrium constant is defined as, the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction.
Given that, $2 X (g) + Y (g)$ \rightleftharpoons 3Z $(g)$
Equilibrium constant $(K_{c}) = \frac{[Product]}{[Reactant]}$
$K_{c} = \frac{[Z]^{3}}{[X]^{2} [Y]}$

Shift-I

Chemical Equilibrium

229244 For the reaction taking place at certain temperature
${NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g)$ ,
if equilibrium pressure is $3 X$ bar then $Δ_{r} G^{0}$ would be

1

- RT \ln 9 - 3 RT \ln X

2

RT \ln 4 - 3 RT \ln X

3

- 3 RT \ln 4 X

4 None of these

Explanation:

From the relation b/w standard Gibb's free energy and equilibrium constant-
$Δ G^{o} = - RT \ln K_{p}; K_{p} = (2 X)^{2} X = 4 X^{3}$
$∴ Δ G^{o} = - RT \ln (4 X^{3})$
$Δ G^{o} = - RT \ln 4 - 3 RT \ln X$

BITSAT 2018

Chemical Equilibrium

229252 The value of $K_{c}$ for the dissociation reaction
$H_{2} (g) ⇌ 2 H (g) is 2.1 \times 10^{- 42}$
This equilibrium mixture contains mainly

1

H_{2} (g)

2

H (g)

atom

3

1 : 1

molar mixture of

H_{2}

and

H

4 1:2 molar mixture of

H_{2}

and

H

Explanation:

For the dissociation reaction -
$H_{2} (g) ⇌ 2 H (g) K_{c} = 2.1 \times 10^{- 42}$
Since, $K_{c}$ is very small.
Hence, $[H_{2}]$ is predominant.

UPTU/UPSEE-2015

Chemical Equilibrium

229255 If $K_{c}$ is the equilibrium constant for the formation of ${NH}_{3}$ , the dissociation constant of ${NH}_{3}$ under the same condition will be

1

\frac{1}{K_{c}}

2

K_{c}^{2}

3

\sqrt{K_{c}}

4

K_{c}

Explanation:

The equilibrium reaction is-
$N_{2} + 3 H_{2} \underset{k_{c}}{\overset{K_{c}}{⇌}} 2 {NH}_{3}$
If the equation representing equilibrium constant is reversed. The new equilibrium constant is inverse of the first value.
$i.e. K_{c}^{'} = \frac{1}{K_{c}}$

CG PET- 2015

Chemical Equilibrium

229234 Identify the correct expression for the equilibrium constant of the following reaction:
$2 X (g) + Y (g) ⇌ 3 Z (g)$

1

K = \frac{[X]^{2} [Y]}{[Z]^{3}}

2

K = \frac{[Z]^{3}}{[X]^{2} [Y]}

3

K = \frac{3 [Z]}{2 [X] [Y]}

4

K = [Z]^{3} [X]^{2} [Y]

Explanation:

Equilibrium constant is defined as, the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction.
Given that, $2 X (g) + Y (g)$ \rightleftharpoons 3Z $(g)$
Equilibrium constant $(K_{c}) = \frac{[Product]}{[Reactant]}$
$K_{c} = \frac{[Z]^{3}}{[X]^{2} [Y]}$

Shift-I

Chemical Equilibrium

229244 For the reaction taking place at certain temperature
${NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g)$ ,
if equilibrium pressure is $3 X$ bar then $Δ_{r} G^{0}$ would be

1

- RT \ln 9 - 3 RT \ln X

2

RT \ln 4 - 3 RT \ln X

3

- 3 RT \ln 4 X

4 None of these

Explanation:

From the relation b/w standard Gibb's free energy and equilibrium constant-
$Δ G^{o} = - RT \ln K_{p}; K_{p} = (2 X)^{2} X = 4 X^{3}$
$∴ Δ G^{o} = - RT \ln (4 X^{3})$
$Δ G^{o} = - RT \ln 4 - 3 RT \ln X$

BITSAT 2018

Chemical Equilibrium

229252 The value of $K_{c}$ for the dissociation reaction
$H_{2} (g) ⇌ 2 H (g) is 2.1 \times 10^{- 42}$
This equilibrium mixture contains mainly

1

H_{2} (g)

2

H (g)

atom

3

1 : 1

molar mixture of

H_{2}

and

H

4 1:2 molar mixture of

H_{2}

and

H

Explanation:

For the dissociation reaction -
$H_{2} (g) ⇌ 2 H (g) K_{c} = 2.1 \times 10^{- 42}$
Since, $K_{c}$ is very small.
Hence, $[H_{2}]$ is predominant.

UPTU/UPSEE-2015

Chemical Equilibrium

229255 If $K_{c}$ is the equilibrium constant for the formation of ${NH}_{3}$ , the dissociation constant of ${NH}_{3}$ under the same condition will be

1

\frac{1}{K_{c}}

2

K_{c}^{2}

3

\sqrt{K_{c}}

4

K_{c}

Explanation:

The equilibrium reaction is-
$N_{2} + 3 H_{2} \underset{k_{c}}{\overset{K_{c}}{⇌}} 2 {NH}_{3}$
If the equation representing equilibrium constant is reversed. The new equilibrium constant is inverse of the first value.
$i.e. K_{c}^{'} = \frac{1}{K_{c}}$

CG PET- 2015

Chemical Equilibrium

229234 Identify the correct expression for the equilibrium constant of the following reaction:
$2 X (g) + Y (g) ⇌ 3 Z (g)$

1

K = \frac{[X]^{2} [Y]}{[Z]^{3}}

2

K = \frac{[Z]^{3}}{[X]^{2} [Y]}

3

K = \frac{3 [Z]}{2 [X] [Y]}

4

K = [Z]^{3} [X]^{2} [Y]

Explanation:

Equilibrium constant is defined as, the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction.
Given that, $2 X (g) + Y (g)$ \rightleftharpoons 3Z $(g)$
Equilibrium constant $(K_{c}) = \frac{[Product]}{[Reactant]}$
$K_{c} = \frac{[Z]^{3}}{[X]^{2} [Y]}$

Shift-I

Chemical Equilibrium

229244 For the reaction taking place at certain temperature
${NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g)$ ,
if equilibrium pressure is $3 X$ bar then $Δ_{r} G^{0}$ would be

1

- RT \ln 9 - 3 RT \ln X

2

RT \ln 4 - 3 RT \ln X

3

- 3 RT \ln 4 X

4 None of these

Explanation:

From the relation b/w standard Gibb's free energy and equilibrium constant-
$Δ G^{o} = - RT \ln K_{p}; K_{p} = (2 X)^{2} X = 4 X^{3}$
$∴ Δ G^{o} = - RT \ln (4 X^{3})$
$Δ G^{o} = - RT \ln 4 - 3 RT \ln X$

BITSAT 2018

Chemical Equilibrium

229252 The value of $K_{c}$ for the dissociation reaction
$H_{2} (g) ⇌ 2 H (g) is 2.1 \times 10^{- 42}$
This equilibrium mixture contains mainly

1

H_{2} (g)

2

H (g)

atom

3

1 : 1

molar mixture of

H_{2}

and

H

4 1:2 molar mixture of

H_{2}

and

H

Explanation:

For the dissociation reaction -
$H_{2} (g) ⇌ 2 H (g) K_{c} = 2.1 \times 10^{- 42}$
Since, $K_{c}$ is very small.
Hence, $[H_{2}]$ is predominant.

UPTU/UPSEE-2015

Chemical Equilibrium

229255 If $K_{c}$ is the equilibrium constant for the formation of ${NH}_{3}$ , the dissociation constant of ${NH}_{3}$ under the same condition will be

1

\frac{1}{K_{c}}

2

K_{c}^{2}

3

\sqrt{K_{c}}

4

K_{c}

Explanation:

The equilibrium reaction is-
$N_{2} + 3 H_{2} \underset{k_{c}}{\overset{K_{c}}{⇌}} 2 {NH}_{3}$
If the equation representing equilibrium constant is reversed. The new equilibrium constant is inverse of the first value.
$i.e. K_{c}^{'} = \frac{1}{K_{c}}$

CG PET- 2015

Question Bank Series

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