QBManager

Chemical Equilibrium

228934 The equilibrium constant $(K_{p})$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{- 4}$ and at $127^{\circ} C$ is $0.60 \times 10^{- 4}$ . Calculate the mean heat of formation of ammonia mole in this temperature range.

1

- 82.64 Cal

2

- 826.4 Cal

3

- 1652.8 Cal

4

- 165.2 Cal

Explanation:

Given that,
$\begin{aligned} K_{1} = 1.2 \times 10^{- 4}, K_{2} = 0.60 \times 10^{- 4} \\ T_{1} = 27^{0} C = 300 K, T_{2} = 127^{0} C = 400 K \end{aligned}$
From van't Hoff's equation -
$\begin{aligned} \log \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ or \log \frac{0.60 \times 10^{- 4}}{1.2 \times 10^{- 4}} = \frac{- Δ H}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{400}) \\ or \log 0.5 = \frac{- Δ H}{19.147} (\frac{100}{300 \times 400}) \\ or Δ H = - 0.3010 \times 19.147 \times 300 \times 4 \\ = - 6915.8964 J \\ = - 1654.52 Cal \end{aligned}$
Hence, the mean heat of formation of ammonia mole $= \frac{- 1654.52}{2} = 827.26 \approx - 826.4 Cal$

Shift-I

Chemical Equilibrium

228935 For the following three reactions (i), (ii) and (iii), equilibrium constants are given
$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1}$
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + H_{2} (g); K_{2}$
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3}$
Which of the following relations is correct?

1

K_{1} \sqrt{K_{2}} = K_{3}

2

K_{1} K_{3} = K_{2}

3

K_{3} = K_{1} K_{2}

4

K_{3} = {(K_{2})}^{3} {(K_{1})}^{2}

Explanation:

$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1} \dots$ (i)
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + 3 H_{2} (g); K_{2} \dots .$ . (ii)
After adding equation (i) and (ii), we get-
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3} \dots .$ . (iii)
Which is equal to equation (iii), so, $K_{3} = K_{1} K_{2}$

Shift-I

Chemical Equilibrium

228941 The equilibrium constant for the following reactions are given at $25^{\circ} C$
$2 A ⇌ B + C, K_{1} = 1.0$
$2 B ⇌ C + D, K_{2} = 16$
$2 C + D ⇌ 2 P, K_{3} = 25$
The equilibrium constant for the reaction
$P ⇌ A + \frac{1}{2} B$ at $25^{\circ} C$ is

1

\frac{1}{20}

2 20

3

\frac{1}{42}

4 21

Explanation:

: Given that,
$\begin{array}{ll} 2 A ◻ B + C, & K_{1} = 10 \\ K_{1} = \frac{[B] [C]}{[A]^{2}} \\ 2 B ◻ C + D, & K_{2} = 16 \\ K_{2} = \frac{[C] [D]}{[B]^{2}} \\ 2 C + D ◻ 2 P, & K_{3} = 25 \\ K_{3} = \frac{[P]^{2}}{[C]^{2} [D]} \end{array}$
Required reaction is:
$\begin{aligned} P ◻ A + \frac{1}{2} B \\ K_{4} = \frac{[A] [B]^{1 / 2}}{[P]} \end{aligned}$
$\begin{aligned} ∴ K_{4} = \sqrt{\frac{1}{K_{3}} \times \frac{1}{K_{2}} \times \frac{1}{K_{1}}} \\ K_{4} = \sqrt{\frac{1}{25} \times \frac{1}{16} \times \frac{1}{1.0}} \\ K_{4} = \frac{1}{5} \times \frac{1}{4} \times 1 \\ K_{4} = \frac{1}{20} \end{aligned}$

WB-JEE-2020

Chemical Equilibrium

228943 Assertion: If value of equilibrium constant is $K_{c}$ for $A + B ⇌ C$ . If chemical reaction is reversed $C ⇌ A + B$ , then value of equilibrium constant will be $1 / K_{c}$ .
Reason: Equilibrium constant depends upon the expression of chemical reaction.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

Given the equilibrium constant $(K_{c})$ of reaction,
$\begin{aligned} A + B ⇌ C, K_{c} \\ C ⇌ A + B, \frac{1}{K_{c}} \end{aligned}$
The value of equilibrium constant depend upon the expression are chemical reaction.
The value of equilibrium constant depend on the stoichiometry equation.

AIIMS 25 May 2019 (Evening)

Chemical Equilibrium

228944 2 moles ${PCl}_{5}$ were introduced in a $2 L$ flask and heated at $625 K$ to establish equilibrium when $60 %$ of ${PCl}_{5}$ was dissociated into ${PCl}_{3}$ and ${Cl}_{2}$ . The equilibrium constant is

1 0.90

2 1.8

3 0.128

4 0.53

Explanation:

The given reaction is $- {PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ Initial conc. $1 0 0$
At equi. $(1 - 0.6) 1 0.6 0.6$
$[∵$ Initial concentration of ${PCl}_{5} = \frac{2 mole}{2 L} = 1 mole / L]$
Now, $K = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}$
$= \frac{0.6 \times 0.6}{0.4} = 0.09 \times 10 = 0.9$

AMU-2019

Chemical Equilibrium

228934 The equilibrium constant $(K_{p})$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{- 4}$ and at $127^{\circ} C$ is $0.60 \times 10^{- 4}$ . Calculate the mean heat of formation of ammonia mole in this temperature range.

1

- 82.64 Cal

2

- 826.4 Cal

3

- 1652.8 Cal

4

- 165.2 Cal

Explanation:

Given that,
$\begin{aligned} K_{1} = 1.2 \times 10^{- 4}, K_{2} = 0.60 \times 10^{- 4} \\ T_{1} = 27^{0} C = 300 K, T_{2} = 127^{0} C = 400 K \end{aligned}$
From van't Hoff's equation -
$\begin{aligned} \log \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ or \log \frac{0.60 \times 10^{- 4}}{1.2 \times 10^{- 4}} = \frac{- Δ H}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{400}) \\ or \log 0.5 = \frac{- Δ H}{19.147} (\frac{100}{300 \times 400}) \\ or Δ H = - 0.3010 \times 19.147 \times 300 \times 4 \\ = - 6915.8964 J \\ = - 1654.52 Cal \end{aligned}$
Hence, the mean heat of formation of ammonia mole $= \frac{- 1654.52}{2} = 827.26 \approx - 826.4 Cal$

Shift-I

Chemical Equilibrium

228935 For the following three reactions (i), (ii) and (iii), equilibrium constants are given
$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1}$
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + H_{2} (g); K_{2}$
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3}$
Which of the following relations is correct?

1

K_{1} \sqrt{K_{2}} = K_{3}

2

K_{1} K_{3} = K_{2}

3

K_{3} = K_{1} K_{2}

4

K_{3} = {(K_{2})}^{3} {(K_{1})}^{2}

Explanation:

$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1} \dots$ (i)
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + 3 H_{2} (g); K_{2} \dots .$ . (ii)
After adding equation (i) and (ii), we get-
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3} \dots .$ . (iii)
Which is equal to equation (iii), so, $K_{3} = K_{1} K_{2}$

Shift-I

Chemical Equilibrium

228941 The equilibrium constant for the following reactions are given at $25^{\circ} C$
$2 A ⇌ B + C, K_{1} = 1.0$
$2 B ⇌ C + D, K_{2} = 16$
$2 C + D ⇌ 2 P, K_{3} = 25$
The equilibrium constant for the reaction
$P ⇌ A + \frac{1}{2} B$ at $25^{\circ} C$ is

1

\frac{1}{20}

2 20

3

\frac{1}{42}

4 21

Explanation:

: Given that,
$\begin{array}{ll} 2 A ◻ B + C, & K_{1} = 10 \\ K_{1} = \frac{[B] [C]}{[A]^{2}} \\ 2 B ◻ C + D, & K_{2} = 16 \\ K_{2} = \frac{[C] [D]}{[B]^{2}} \\ 2 C + D ◻ 2 P, & K_{3} = 25 \\ K_{3} = \frac{[P]^{2}}{[C]^{2} [D]} \end{array}$
Required reaction is:
$\begin{aligned} P ◻ A + \frac{1}{2} B \\ K_{4} = \frac{[A] [B]^{1 / 2}}{[P]} \end{aligned}$
$\begin{aligned} ∴ K_{4} = \sqrt{\frac{1}{K_{3}} \times \frac{1}{K_{2}} \times \frac{1}{K_{1}}} \\ K_{4} = \sqrt{\frac{1}{25} \times \frac{1}{16} \times \frac{1}{1.0}} \\ K_{4} = \frac{1}{5} \times \frac{1}{4} \times 1 \\ K_{4} = \frac{1}{20} \end{aligned}$

WB-JEE-2020

Chemical Equilibrium

228943 Assertion: If value of equilibrium constant is $K_{c}$ for $A + B ⇌ C$ . If chemical reaction is reversed $C ⇌ A + B$ , then value of equilibrium constant will be $1 / K_{c}$ .
Reason: Equilibrium constant depends upon the expression of chemical reaction.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

Given the equilibrium constant $(K_{c})$ of reaction,
$\begin{aligned} A + B ⇌ C, K_{c} \\ C ⇌ A + B, \frac{1}{K_{c}} \end{aligned}$
The value of equilibrium constant depend upon the expression are chemical reaction.
The value of equilibrium constant depend on the stoichiometry equation.

AIIMS 25 May 2019 (Evening)

Chemical Equilibrium

228944 2 moles ${PCl}_{5}$ were introduced in a $2 L$ flask and heated at $625 K$ to establish equilibrium when $60 %$ of ${PCl}_{5}$ was dissociated into ${PCl}_{3}$ and ${Cl}_{2}$ . The equilibrium constant is

1 0.90

2 1.8

3 0.128

4 0.53

Explanation:

The given reaction is $- {PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ Initial conc. $1 0 0$
At equi. $(1 - 0.6) 1 0.6 0.6$
$[∵$ Initial concentration of ${PCl}_{5} = \frac{2 mole}{2 L} = 1 mole / L]$
Now, $K = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}$
$= \frac{0.6 \times 0.6}{0.4} = 0.09 \times 10 = 0.9$

AMU-2019

Chemical Equilibrium

228934 The equilibrium constant $(K_{p})$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{- 4}$ and at $127^{\circ} C$ is $0.60 \times 10^{- 4}$ . Calculate the mean heat of formation of ammonia mole in this temperature range.

1

- 82.64 Cal

2

- 826.4 Cal

3

- 1652.8 Cal

4

- 165.2 Cal

Explanation:

Given that,
$\begin{aligned} K_{1} = 1.2 \times 10^{- 4}, K_{2} = 0.60 \times 10^{- 4} \\ T_{1} = 27^{0} C = 300 K, T_{2} = 127^{0} C = 400 K \end{aligned}$
From van't Hoff's equation -
$\begin{aligned} \log \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ or \log \frac{0.60 \times 10^{- 4}}{1.2 \times 10^{- 4}} = \frac{- Δ H}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{400}) \\ or \log 0.5 = \frac{- Δ H}{19.147} (\frac{100}{300 \times 400}) \\ or Δ H = - 0.3010 \times 19.147 \times 300 \times 4 \\ = - 6915.8964 J \\ = - 1654.52 Cal \end{aligned}$
Hence, the mean heat of formation of ammonia mole $= \frac{- 1654.52}{2} = 827.26 \approx - 826.4 Cal$

Shift-I

Chemical Equilibrium

228935 For the following three reactions (i), (ii) and (iii), equilibrium constants are given
$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1}$
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + H_{2} (g); K_{2}$
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3}$
Which of the following relations is correct?

1

K_{1} \sqrt{K_{2}} = K_{3}

2

K_{1} K_{3} = K_{2}

3

K_{3} = K_{1} K_{2}

4

K_{3} = {(K_{2})}^{3} {(K_{1})}^{2}

Explanation:

$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1} \dots$ (i)
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + 3 H_{2} (g); K_{2} \dots .$ . (ii)
After adding equation (i) and (ii), we get-
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3} \dots .$ . (iii)
Which is equal to equation (iii), so, $K_{3} = K_{1} K_{2}$

Shift-I

Chemical Equilibrium

228941 The equilibrium constant for the following reactions are given at $25^{\circ} C$
$2 A ⇌ B + C, K_{1} = 1.0$
$2 B ⇌ C + D, K_{2} = 16$
$2 C + D ⇌ 2 P, K_{3} = 25$
The equilibrium constant for the reaction
$P ⇌ A + \frac{1}{2} B$ at $25^{\circ} C$ is

1

\frac{1}{20}

2 20

3

\frac{1}{42}

4 21

Explanation:

: Given that,
$\begin{array}{ll} 2 A ◻ B + C, & K_{1} = 10 \\ K_{1} = \frac{[B] [C]}{[A]^{2}} \\ 2 B ◻ C + D, & K_{2} = 16 \\ K_{2} = \frac{[C] [D]}{[B]^{2}} \\ 2 C + D ◻ 2 P, & K_{3} = 25 \\ K_{3} = \frac{[P]^{2}}{[C]^{2} [D]} \end{array}$
Required reaction is:
$\begin{aligned} P ◻ A + \frac{1}{2} B \\ K_{4} = \frac{[A] [B]^{1 / 2}}{[P]} \end{aligned}$
$\begin{aligned} ∴ K_{4} = \sqrt{\frac{1}{K_{3}} \times \frac{1}{K_{2}} \times \frac{1}{K_{1}}} \\ K_{4} = \sqrt{\frac{1}{25} \times \frac{1}{16} \times \frac{1}{1.0}} \\ K_{4} = \frac{1}{5} \times \frac{1}{4} \times 1 \\ K_{4} = \frac{1}{20} \end{aligned}$

WB-JEE-2020

Chemical Equilibrium

228943 Assertion: If value of equilibrium constant is $K_{c}$ for $A + B ⇌ C$ . If chemical reaction is reversed $C ⇌ A + B$ , then value of equilibrium constant will be $1 / K_{c}$ .
Reason: Equilibrium constant depends upon the expression of chemical reaction.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

Given the equilibrium constant $(K_{c})$ of reaction,
$\begin{aligned} A + B ⇌ C, K_{c} \\ C ⇌ A + B, \frac{1}{K_{c}} \end{aligned}$
The value of equilibrium constant depend upon the expression are chemical reaction.
The value of equilibrium constant depend on the stoichiometry equation.

AIIMS 25 May 2019 (Evening)

Chemical Equilibrium

228944 2 moles ${PCl}_{5}$ were introduced in a $2 L$ flask and heated at $625 K$ to establish equilibrium when $60 %$ of ${PCl}_{5}$ was dissociated into ${PCl}_{3}$ and ${Cl}_{2}$ . The equilibrium constant is

1 0.90

2 1.8

3 0.128

4 0.53

Explanation:

The given reaction is $- {PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ Initial conc. $1 0 0$
At equi. $(1 - 0.6) 1 0.6 0.6$
$[∵$ Initial concentration of ${PCl}_{5} = \frac{2 mole}{2 L} = 1 mole / L]$
Now, $K = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}$
$= \frac{0.6 \times 0.6}{0.4} = 0.09 \times 10 = 0.9$

AMU-2019

Chemical Equilibrium

228934 The equilibrium constant $(K_{p})$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{- 4}$ and at $127^{\circ} C$ is $0.60 \times 10^{- 4}$ . Calculate the mean heat of formation of ammonia mole in this temperature range.

1

- 82.64 Cal

2

- 826.4 Cal

3

- 1652.8 Cal

4

- 165.2 Cal

Explanation:

Given that,
$\begin{aligned} K_{1} = 1.2 \times 10^{- 4}, K_{2} = 0.60 \times 10^{- 4} \\ T_{1} = 27^{0} C = 300 K, T_{2} = 127^{0} C = 400 K \end{aligned}$
From van't Hoff's equation -
$\begin{aligned} \log \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ or \log \frac{0.60 \times 10^{- 4}}{1.2 \times 10^{- 4}} = \frac{- Δ H}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{400}) \\ or \log 0.5 = \frac{- Δ H}{19.147} (\frac{100}{300 \times 400}) \\ or Δ H = - 0.3010 \times 19.147 \times 300 \times 4 \\ = - 6915.8964 J \\ = - 1654.52 Cal \end{aligned}$
Hence, the mean heat of formation of ammonia mole $= \frac{- 1654.52}{2} = 827.26 \approx - 826.4 Cal$

Shift-I

Chemical Equilibrium

228935 For the following three reactions (i), (ii) and (iii), equilibrium constants are given
$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1}$
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + H_{2} (g); K_{2}$
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3}$
Which of the following relations is correct?

1

K_{1} \sqrt{K_{2}} = K_{3}

2

K_{1} K_{3} = K_{2}

3

K_{3} = K_{1} K_{2}

4

K_{3} = {(K_{2})}^{3} {(K_{1})}^{2}

Explanation:

$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1} \dots$ (i)
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + 3 H_{2} (g); K_{2} \dots .$ . (ii)
After adding equation (i) and (ii), we get-
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3} \dots .$ . (iii)
Which is equal to equation (iii), so, $K_{3} = K_{1} K_{2}$

Shift-I

Chemical Equilibrium

228941 The equilibrium constant for the following reactions are given at $25^{\circ} C$
$2 A ⇌ B + C, K_{1} = 1.0$
$2 B ⇌ C + D, K_{2} = 16$
$2 C + D ⇌ 2 P, K_{3} = 25$
The equilibrium constant for the reaction
$P ⇌ A + \frac{1}{2} B$ at $25^{\circ} C$ is

1

\frac{1}{20}

2 20

3

\frac{1}{42}

4 21

Explanation:

: Given that,
$\begin{array}{ll} 2 A ◻ B + C, & K_{1} = 10 \\ K_{1} = \frac{[B] [C]}{[A]^{2}} \\ 2 B ◻ C + D, & K_{2} = 16 \\ K_{2} = \frac{[C] [D]}{[B]^{2}} \\ 2 C + D ◻ 2 P, & K_{3} = 25 \\ K_{3} = \frac{[P]^{2}}{[C]^{2} [D]} \end{array}$
Required reaction is:
$\begin{aligned} P ◻ A + \frac{1}{2} B \\ K_{4} = \frac{[A] [B]^{1 / 2}}{[P]} \end{aligned}$
$\begin{aligned} ∴ K_{4} = \sqrt{\frac{1}{K_{3}} \times \frac{1}{K_{2}} \times \frac{1}{K_{1}}} \\ K_{4} = \sqrt{\frac{1}{25} \times \frac{1}{16} \times \frac{1}{1.0}} \\ K_{4} = \frac{1}{5} \times \frac{1}{4} \times 1 \\ K_{4} = \frac{1}{20} \end{aligned}$

WB-JEE-2020

Chemical Equilibrium

228943 Assertion: If value of equilibrium constant is $K_{c}$ for $A + B ⇌ C$ . If chemical reaction is reversed $C ⇌ A + B$ , then value of equilibrium constant will be $1 / K_{c}$ .
Reason: Equilibrium constant depends upon the expression of chemical reaction.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

Given the equilibrium constant $(K_{c})$ of reaction,
$\begin{aligned} A + B ⇌ C, K_{c} \\ C ⇌ A + B, \frac{1}{K_{c}} \end{aligned}$
The value of equilibrium constant depend upon the expression are chemical reaction.
The value of equilibrium constant depend on the stoichiometry equation.

AIIMS 25 May 2019 (Evening)

Chemical Equilibrium

228944 2 moles ${PCl}_{5}$ were introduced in a $2 L$ flask and heated at $625 K$ to establish equilibrium when $60 %$ of ${PCl}_{5}$ was dissociated into ${PCl}_{3}$ and ${Cl}_{2}$ . The equilibrium constant is

1 0.90

2 1.8

3 0.128

4 0.53

Explanation:

The given reaction is $- {PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ Initial conc. $1 0 0$
At equi. $(1 - 0.6) 1 0.6 0.6$
$[∵$ Initial concentration of ${PCl}_{5} = \frac{2 mole}{2 L} = 1 mole / L]$
Now, $K = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}$
$= \frac{0.6 \times 0.6}{0.4} = 0.09 \times 10 = 0.9$

AMU-2019

Chemical Equilibrium

228934 The equilibrium constant $(K_{p})$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{- 4}$ and at $127^{\circ} C$ is $0.60 \times 10^{- 4}$ . Calculate the mean heat of formation of ammonia mole in this temperature range.

1

- 82.64 Cal

2

- 826.4 Cal

3

- 1652.8 Cal

4

- 165.2 Cal

Explanation:

Given that,
$\begin{aligned} K_{1} = 1.2 \times 10^{- 4}, K_{2} = 0.60 \times 10^{- 4} \\ T_{1} = 27^{0} C = 300 K, T_{2} = 127^{0} C = 400 K \end{aligned}$
From van't Hoff's equation -
$\begin{aligned} \log \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ or \log \frac{0.60 \times 10^{- 4}}{1.2 \times 10^{- 4}} = \frac{- Δ H}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{400}) \\ or \log 0.5 = \frac{- Δ H}{19.147} (\frac{100}{300 \times 400}) \\ or Δ H = - 0.3010 \times 19.147 \times 300 \times 4 \\ = - 6915.8964 J \\ = - 1654.52 Cal \end{aligned}$
Hence, the mean heat of formation of ammonia mole $= \frac{- 1654.52}{2} = 827.26 \approx - 826.4 Cal$

Shift-I

Chemical Equilibrium

228935 For the following three reactions (i), (ii) and (iii), equilibrium constants are given
$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1}$
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + H_{2} (g); K_{2}$
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3}$
Which of the following relations is correct?

1

K_{1} \sqrt{K_{2}} = K_{3}

2

K_{1} K_{3} = K_{2}

3

K_{3} = K_{1} K_{2}

4

K_{3} = {(K_{2})}^{3} {(K_{1})}^{2}

Explanation:

$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g); K_{1} \dots$ (i)
${CH}_{4} (g) + H_{2} O (g) ⇌ CO (g) + 3 H_{2} (g); K_{2} \dots .$ . (ii)
After adding equation (i) and (ii), we get-
${CH}_{4} (g) + 2 H_{2} O (g) ⇌ {CO}_{2} (g) + 4 H_{2} (g); K_{3} \dots .$ . (iii)
Which is equal to equation (iii), so, $K_{3} = K_{1} K_{2}$

Shift-I

Chemical Equilibrium

228941 The equilibrium constant for the following reactions are given at $25^{\circ} C$
$2 A ⇌ B + C, K_{1} = 1.0$
$2 B ⇌ C + D, K_{2} = 16$
$2 C + D ⇌ 2 P, K_{3} = 25$
The equilibrium constant for the reaction
$P ⇌ A + \frac{1}{2} B$ at $25^{\circ} C$ is

1

\frac{1}{20}

2 20

3

\frac{1}{42}

4 21

Explanation:

: Given that,
$\begin{array}{ll} 2 A ◻ B + C, & K_{1} = 10 \\ K_{1} = \frac{[B] [C]}{[A]^{2}} \\ 2 B ◻ C + D, & K_{2} = 16 \\ K_{2} = \frac{[C] [D]}{[B]^{2}} \\ 2 C + D ◻ 2 P, & K_{3} = 25 \\ K_{3} = \frac{[P]^{2}}{[C]^{2} [D]} \end{array}$
Required reaction is:
$\begin{aligned} P ◻ A + \frac{1}{2} B \\ K_{4} = \frac{[A] [B]^{1 / 2}}{[P]} \end{aligned}$
$\begin{aligned} ∴ K_{4} = \sqrt{\frac{1}{K_{3}} \times \frac{1}{K_{2}} \times \frac{1}{K_{1}}} \\ K_{4} = \sqrt{\frac{1}{25} \times \frac{1}{16} \times \frac{1}{1.0}} \\ K_{4} = \frac{1}{5} \times \frac{1}{4} \times 1 \\ K_{4} = \frac{1}{20} \end{aligned}$

WB-JEE-2020

Chemical Equilibrium

228943 Assertion: If value of equilibrium constant is $K_{c}$ for $A + B ⇌ C$ . If chemical reaction is reversed $C ⇌ A + B$ , then value of equilibrium constant will be $1 / K_{c}$ .
Reason: Equilibrium constant depends upon the expression of chemical reaction.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

Given the equilibrium constant $(K_{c})$ of reaction,
$\begin{aligned} A + B ⇌ C, K_{c} \\ C ⇌ A + B, \frac{1}{K_{c}} \end{aligned}$
The value of equilibrium constant depend upon the expression are chemical reaction.
The value of equilibrium constant depend on the stoichiometry equation.

AIIMS 25 May 2019 (Evening)

Chemical Equilibrium

228944 2 moles ${PCl}_{5}$ were introduced in a $2 L$ flask and heated at $625 K$ to establish equilibrium when $60 %$ of ${PCl}_{5}$ was dissociated into ${PCl}_{3}$ and ${Cl}_{2}$ . The equilibrium constant is

1 0.90

2 1.8

3 0.128

4 0.53

Explanation:

The given reaction is $- {PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$ Initial conc. $1 0 0$
At equi. $(1 - 0.6) 1 0.6 0.6$
$[∵$ Initial concentration of ${PCl}_{5} = \frac{2 mole}{2 L} = 1 mole / L]$
Now, $K = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}$
$= \frac{0.6 \times 0.6}{0.4} = 0.09 \times 10 = 0.9$

AMU-2019

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