QBManager

Chemical Equilibrium

229284 Calculate $K_{c}$ for the reversible process given below if $K_{p} = 167$ and $T = 800^{\circ} C$ .
${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$

1 1.95

2 1.85

3 1.89

4 1.60

Explanation:

Given, $K_{p} = 167, T = 800^{\circ}, K_{c} =$ ?
The decomposition reaction of lime stone is -
$\begin{array}{ll} ∴ & {CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g) \\ K_{p} = K_{c} (RT)^{Δ n_{g}} \\ Δ n_{g} = 1 - 0 = 1 \\ K_{c} = \frac{K_{p}}{[0.0821 \times (800 + 273)]^{1}} \\ K_{c} = \frac{167}{88.09} \\ K_{c} \approx 1.89 \end{array}$

WB-JEE-2010

Chemical Equilibrium

229287 If the equilibrium constant of the reaction $2 HI ⇌ H_{2} + I_{2}$ is 0.25 , then the equilibrium constant for the reaction $H_{2} + I_{2} ⇌ 2 HI$ would be

1 1

2 2

3 3

4 4

Explanation:

$2 HI ⇌ H_{2} + I_{2}$ For this chemical reaction equilibrium constant -
$K_{c} = 0.25$
Now the reaction is reveres
$H_{2} + I_{2} ⇌ 2 HI$
Hence, new equilibrium constant is -
$\begin{aligned} K_{c}^{'} & = \frac{1}{K_{c}} \\ = \frac{1}{0.25} \\ = \frac{100}{25} \\ K_{c}^{'} & = 4 \end{aligned}$

BITSAT 2009

Chemical Equilibrium

229288 For the reaction, $2 B + A ⇌ C$ , the equilibrium constant is-

1

\frac{[A] [B]^{3}}{[C]}

2

\frac{[C]}{[A] [2 B]}

3

\frac{[C]}{[A] [B]^{2}}

4

\frac{[A] [B]}{[C]}

Explanation:

We know equilibrium constant of any reaction is -
$\begin{array}{ll} K_{c} = \frac{[Product]}{[Reactent]} \\ So, \\ 2 B + A ⇌ C \\ K_{c} = \frac{[C]}{[B]^{2} [A]} \end{array}$

BCECE-2009

Chemical Equilibrium

229289 One mole of $N_{2} O_{4}$ is heated in a flask with a volume of $0.1 {dm}^{3}$ . At equilibrium 1.708 mole of ${NO}_{2}$ and 0.146 mole of $N_{2} O_{4}$ were found at $134^{0} C$ . The equilibrium constant will be

1

250 mol {dm}^{- 3}

2

300 mol {dm}^{- 3}

3

200 mol {dm}^{- 3}

4

230 mol {dm}^{- 3}

Explanation:

Given data :
Initially-
Mole of $N_{2} O_{4} = 1$
Volume $= 0.1 {dm}^{3}$
At equilibrium-
Moles of ${NO}_{2} = 1.708$
Moles of $N_{2} O_{4} = 0.146$
For the equilibrium reaction-
$N_{2} O_{4} ⇌ 2 {NO}_{2}$
$\begin{aligned} K_{c} & = \frac{{[{NO}_{2}]}^{2}}{[N_{2} O_{4}]} \\ K_{c} & = \frac{{(1.708 mole / 0.1 {dm}^{3})}^{2}}{0.146 mole / 0.1 {dm}^{3}} \\ K_{c} & = 200 mol {dm}^{- 3} \end{aligned}$

MPPET-2009

Chemical Equilibrium

229290 Assertion : For reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Unit of $K_{c} = L^{2} {mol}^{- 2}$
Reason : For the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Equilibrium constant, $K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

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

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

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

For the reaction -
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
$K_{C} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{[{molL}^{- 1}]}^{2}}{[{molL}^{- 1}] {[mol L^{- 1}]}^{3}}$
$= {mol}^{- 1} L^{2}$

AIIMS-(2008)

Chemical Equilibrium

229284 Calculate $K_{c}$ for the reversible process given below if $K_{p} = 167$ and $T = 800^{\circ} C$ .
${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$

1 1.95

2 1.85

3 1.89

4 1.60

Explanation:

Given, $K_{p} = 167, T = 800^{\circ}, K_{c} =$ ?
The decomposition reaction of lime stone is -
$\begin{array}{ll} ∴ & {CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g) \\ K_{p} = K_{c} (RT)^{Δ n_{g}} \\ Δ n_{g} = 1 - 0 = 1 \\ K_{c} = \frac{K_{p}}{[0.0821 \times (800 + 273)]^{1}} \\ K_{c} = \frac{167}{88.09} \\ K_{c} \approx 1.89 \end{array}$

WB-JEE-2010

Chemical Equilibrium

229287 If the equilibrium constant of the reaction $2 HI ⇌ H_{2} + I_{2}$ is 0.25 , then the equilibrium constant for the reaction $H_{2} + I_{2} ⇌ 2 HI$ would be

1 1

2 2

3 3

4 4

Explanation:

$2 HI ⇌ H_{2} + I_{2}$ For this chemical reaction equilibrium constant -
$K_{c} = 0.25$
Now the reaction is reveres
$H_{2} + I_{2} ⇌ 2 HI$
Hence, new equilibrium constant is -
$\begin{aligned} K_{c}^{'} & = \frac{1}{K_{c}} \\ = \frac{1}{0.25} \\ = \frac{100}{25} \\ K_{c}^{'} & = 4 \end{aligned}$

BITSAT 2009

Chemical Equilibrium

229288 For the reaction, $2 B + A ⇌ C$ , the equilibrium constant is-

1

\frac{[A] [B]^{3}}{[C]}

2

\frac{[C]}{[A] [2 B]}

3

\frac{[C]}{[A] [B]^{2}}

4

\frac{[A] [B]}{[C]}

Explanation:

We know equilibrium constant of any reaction is -
$\begin{array}{ll} K_{c} = \frac{[Product]}{[Reactent]} \\ So, \\ 2 B + A ⇌ C \\ K_{c} = \frac{[C]}{[B]^{2} [A]} \end{array}$

BCECE-2009

Chemical Equilibrium

229289 One mole of $N_{2} O_{4}$ is heated in a flask with a volume of $0.1 {dm}^{3}$ . At equilibrium 1.708 mole of ${NO}_{2}$ and 0.146 mole of $N_{2} O_{4}$ were found at $134^{0} C$ . The equilibrium constant will be

1

250 mol {dm}^{- 3}

2

300 mol {dm}^{- 3}

3

200 mol {dm}^{- 3}

4

230 mol {dm}^{- 3}

Explanation:

Given data :
Initially-
Mole of $N_{2} O_{4} = 1$
Volume $= 0.1 {dm}^{3}$
At equilibrium-
Moles of ${NO}_{2} = 1.708$
Moles of $N_{2} O_{4} = 0.146$
For the equilibrium reaction-
$N_{2} O_{4} ⇌ 2 {NO}_{2}$
$\begin{aligned} K_{c} & = \frac{{[{NO}_{2}]}^{2}}{[N_{2} O_{4}]} \\ K_{c} & = \frac{{(1.708 mole / 0.1 {dm}^{3})}^{2}}{0.146 mole / 0.1 {dm}^{3}} \\ K_{c} & = 200 mol {dm}^{- 3} \end{aligned}$

MPPET-2009

Chemical Equilibrium

229290 Assertion : For reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Unit of $K_{c} = L^{2} {mol}^{- 2}$
Reason : For the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Equilibrium constant, $K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

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

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

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

For the reaction -
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
$K_{C} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{[{molL}^{- 1}]}^{2}}{[{molL}^{- 1}] {[mol L^{- 1}]}^{3}}$
$= {mol}^{- 1} L^{2}$

AIIMS-(2008)

Chemical Equilibrium

229284 Calculate $K_{c}$ for the reversible process given below if $K_{p} = 167$ and $T = 800^{\circ} C$ .
${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$

1 1.95

2 1.85

3 1.89

4 1.60

Explanation:

Given, $K_{p} = 167, T = 800^{\circ}, K_{c} =$ ?
The decomposition reaction of lime stone is -
$\begin{array}{ll} ∴ & {CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g) \\ K_{p} = K_{c} (RT)^{Δ n_{g}} \\ Δ n_{g} = 1 - 0 = 1 \\ K_{c} = \frac{K_{p}}{[0.0821 \times (800 + 273)]^{1}} \\ K_{c} = \frac{167}{88.09} \\ K_{c} \approx 1.89 \end{array}$

WB-JEE-2010

Chemical Equilibrium

229287 If the equilibrium constant of the reaction $2 HI ⇌ H_{2} + I_{2}$ is 0.25 , then the equilibrium constant for the reaction $H_{2} + I_{2} ⇌ 2 HI$ would be

1 1

2 2

3 3

4 4

Explanation:

$2 HI ⇌ H_{2} + I_{2}$ For this chemical reaction equilibrium constant -
$K_{c} = 0.25$
Now the reaction is reveres
$H_{2} + I_{2} ⇌ 2 HI$
Hence, new equilibrium constant is -
$\begin{aligned} K_{c}^{'} & = \frac{1}{K_{c}} \\ = \frac{1}{0.25} \\ = \frac{100}{25} \\ K_{c}^{'} & = 4 \end{aligned}$

BITSAT 2009

Chemical Equilibrium

229288 For the reaction, $2 B + A ⇌ C$ , the equilibrium constant is-

1

\frac{[A] [B]^{3}}{[C]}

2

\frac{[C]}{[A] [2 B]}

3

\frac{[C]}{[A] [B]^{2}}

4

\frac{[A] [B]}{[C]}

Explanation:

We know equilibrium constant of any reaction is -
$\begin{array}{ll} K_{c} = \frac{[Product]}{[Reactent]} \\ So, \\ 2 B + A ⇌ C \\ K_{c} = \frac{[C]}{[B]^{2} [A]} \end{array}$

BCECE-2009

Chemical Equilibrium

229289 One mole of $N_{2} O_{4}$ is heated in a flask with a volume of $0.1 {dm}^{3}$ . At equilibrium 1.708 mole of ${NO}_{2}$ and 0.146 mole of $N_{2} O_{4}$ were found at $134^{0} C$ . The equilibrium constant will be

1

250 mol {dm}^{- 3}

2

300 mol {dm}^{- 3}

3

200 mol {dm}^{- 3}

4

230 mol {dm}^{- 3}

Explanation:

Given data :
Initially-
Mole of $N_{2} O_{4} = 1$
Volume $= 0.1 {dm}^{3}$
At equilibrium-
Moles of ${NO}_{2} = 1.708$
Moles of $N_{2} O_{4} = 0.146$
For the equilibrium reaction-
$N_{2} O_{4} ⇌ 2 {NO}_{2}$
$\begin{aligned} K_{c} & = \frac{{[{NO}_{2}]}^{2}}{[N_{2} O_{4}]} \\ K_{c} & = \frac{{(1.708 mole / 0.1 {dm}^{3})}^{2}}{0.146 mole / 0.1 {dm}^{3}} \\ K_{c} & = 200 mol {dm}^{- 3} \end{aligned}$

MPPET-2009

Chemical Equilibrium

229290 Assertion : For reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Unit of $K_{c} = L^{2} {mol}^{- 2}$
Reason : For the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Equilibrium constant, $K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

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

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

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

For the reaction -
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
$K_{C} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{[{molL}^{- 1}]}^{2}}{[{molL}^{- 1}] {[mol L^{- 1}]}^{3}}$
$= {mol}^{- 1} L^{2}$

AIIMS-(2008)

Chemical Equilibrium

229284 Calculate $K_{c}$ for the reversible process given below if $K_{p} = 167$ and $T = 800^{\circ} C$ .
${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$

1 1.95

2 1.85

3 1.89

4 1.60

Explanation:

Given, $K_{p} = 167, T = 800^{\circ}, K_{c} =$ ?
The decomposition reaction of lime stone is -
$\begin{array}{ll} ∴ & {CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g) \\ K_{p} = K_{c} (RT)^{Δ n_{g}} \\ Δ n_{g} = 1 - 0 = 1 \\ K_{c} = \frac{K_{p}}{[0.0821 \times (800 + 273)]^{1}} \\ K_{c} = \frac{167}{88.09} \\ K_{c} \approx 1.89 \end{array}$

WB-JEE-2010

Chemical Equilibrium

229287 If the equilibrium constant of the reaction $2 HI ⇌ H_{2} + I_{2}$ is 0.25 , then the equilibrium constant for the reaction $H_{2} + I_{2} ⇌ 2 HI$ would be

1 1

2 2

3 3

4 4

Explanation:

$2 HI ⇌ H_{2} + I_{2}$ For this chemical reaction equilibrium constant -
$K_{c} = 0.25$
Now the reaction is reveres
$H_{2} + I_{2} ⇌ 2 HI$
Hence, new equilibrium constant is -
$\begin{aligned} K_{c}^{'} & = \frac{1}{K_{c}} \\ = \frac{1}{0.25} \\ = \frac{100}{25} \\ K_{c}^{'} & = 4 \end{aligned}$

BITSAT 2009

Chemical Equilibrium

229288 For the reaction, $2 B + A ⇌ C$ , the equilibrium constant is-

1

\frac{[A] [B]^{3}}{[C]}

2

\frac{[C]}{[A] [2 B]}

3

\frac{[C]}{[A] [B]^{2}}

4

\frac{[A] [B]}{[C]}

Explanation:

We know equilibrium constant of any reaction is -
$\begin{array}{ll} K_{c} = \frac{[Product]}{[Reactent]} \\ So, \\ 2 B + A ⇌ C \\ K_{c} = \frac{[C]}{[B]^{2} [A]} \end{array}$

BCECE-2009

Chemical Equilibrium

229289 One mole of $N_{2} O_{4}$ is heated in a flask with a volume of $0.1 {dm}^{3}$ . At equilibrium 1.708 mole of ${NO}_{2}$ and 0.146 mole of $N_{2} O_{4}$ were found at $134^{0} C$ . The equilibrium constant will be

1

250 mol {dm}^{- 3}

2

300 mol {dm}^{- 3}

3

200 mol {dm}^{- 3}

4

230 mol {dm}^{- 3}

Explanation:

Given data :
Initially-
Mole of $N_{2} O_{4} = 1$
Volume $= 0.1 {dm}^{3}$
At equilibrium-
Moles of ${NO}_{2} = 1.708$
Moles of $N_{2} O_{4} = 0.146$
For the equilibrium reaction-
$N_{2} O_{4} ⇌ 2 {NO}_{2}$
$\begin{aligned} K_{c} & = \frac{{[{NO}_{2}]}^{2}}{[N_{2} O_{4}]} \\ K_{c} & = \frac{{(1.708 mole / 0.1 {dm}^{3})}^{2}}{0.146 mole / 0.1 {dm}^{3}} \\ K_{c} & = 200 mol {dm}^{- 3} \end{aligned}$

MPPET-2009

Chemical Equilibrium

229290 Assertion : For reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Unit of $K_{c} = L^{2} {mol}^{- 2}$
Reason : For the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Equilibrium constant, $K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

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

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

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

For the reaction -
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
$K_{C} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{[{molL}^{- 1}]}^{2}}{[{molL}^{- 1}] {[mol L^{- 1}]}^{3}}$
$= {mol}^{- 1} L^{2}$

AIIMS-(2008)

Chemical Equilibrium

229284 Calculate $K_{c}$ for the reversible process given below if $K_{p} = 167$ and $T = 800^{\circ} C$ .
${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$

1 1.95

2 1.85

3 1.89

4 1.60

Explanation:

Given, $K_{p} = 167, T = 800^{\circ}, K_{c} =$ ?
The decomposition reaction of lime stone is -
$\begin{array}{ll} ∴ & {CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g) \\ K_{p} = K_{c} (RT)^{Δ n_{g}} \\ Δ n_{g} = 1 - 0 = 1 \\ K_{c} = \frac{K_{p}}{[0.0821 \times (800 + 273)]^{1}} \\ K_{c} = \frac{167}{88.09} \\ K_{c} \approx 1.89 \end{array}$

WB-JEE-2010

Chemical Equilibrium

229287 If the equilibrium constant of the reaction $2 HI ⇌ H_{2} + I_{2}$ is 0.25 , then the equilibrium constant for the reaction $H_{2} + I_{2} ⇌ 2 HI$ would be

1 1

2 2

3 3

4 4

Explanation:

$2 HI ⇌ H_{2} + I_{2}$ For this chemical reaction equilibrium constant -
$K_{c} = 0.25$
Now the reaction is reveres
$H_{2} + I_{2} ⇌ 2 HI$
Hence, new equilibrium constant is -
$\begin{aligned} K_{c}^{'} & = \frac{1}{K_{c}} \\ = \frac{1}{0.25} \\ = \frac{100}{25} \\ K_{c}^{'} & = 4 \end{aligned}$

BITSAT 2009

Chemical Equilibrium

229288 For the reaction, $2 B + A ⇌ C$ , the equilibrium constant is-

1

\frac{[A] [B]^{3}}{[C]}

2

\frac{[C]}{[A] [2 B]}

3

\frac{[C]}{[A] [B]^{2}}

4

\frac{[A] [B]}{[C]}

Explanation:

We know equilibrium constant of any reaction is -
$\begin{array}{ll} K_{c} = \frac{[Product]}{[Reactent]} \\ So, \\ 2 B + A ⇌ C \\ K_{c} = \frac{[C]}{[B]^{2} [A]} \end{array}$

BCECE-2009

Chemical Equilibrium

229289 One mole of $N_{2} O_{4}$ is heated in a flask with a volume of $0.1 {dm}^{3}$ . At equilibrium 1.708 mole of ${NO}_{2}$ and 0.146 mole of $N_{2} O_{4}$ were found at $134^{0} C$ . The equilibrium constant will be

1

250 mol {dm}^{- 3}

2

300 mol {dm}^{- 3}

3

200 mol {dm}^{- 3}

4

230 mol {dm}^{- 3}

Explanation:

Given data :
Initially-
Mole of $N_{2} O_{4} = 1$
Volume $= 0.1 {dm}^{3}$
At equilibrium-
Moles of ${NO}_{2} = 1.708$
Moles of $N_{2} O_{4} = 0.146$
For the equilibrium reaction-
$N_{2} O_{4} ⇌ 2 {NO}_{2}$
$\begin{aligned} K_{c} & = \frac{{[{NO}_{2}]}^{2}}{[N_{2} O_{4}]} \\ K_{c} & = \frac{{(1.708 mole / 0.1 {dm}^{3})}^{2}}{0.146 mole / 0.1 {dm}^{3}} \\ K_{c} & = 200 mol {dm}^{- 3} \end{aligned}$

MPPET-2009

Chemical Equilibrium

229290 Assertion : For reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Unit of $K_{c} = L^{2} {mol}^{- 2}$
Reason : For the reaction
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
Equilibrium constant, $K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

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

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

3 In the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the assertion is incorrect but the Reason is correct.

Explanation:

For the reaction -
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$
$K_{C} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{[{molL}^{- 1}]}^{2}}{[{molL}^{- 1}] {[mol L^{- 1}]}^{3}}$
$= {mol}^{- 1} L^{2}$

AIIMS-(2008)