QBManager

Chemical Equilibrium

229257 What will be the value of ${[{OH}^{-}]}^{2}$ in the $0.1 M$ solution of ammonium hydroxide having $K_{b} =$ $1.8 \times 10^{- 5}$ ?

1

1.8 \times 10^{- 7}

2

1.8 \times 10^{- 6}

3

1.8 \times 10^{- 4}

4

1.8 \times 10^{- 3}

Explanation:

The dissociation reaction of ${NH}_{4} OH$ is -
$\begin{array}{cccc} {NH}_{4} OH & ⇌ & {NH}_{4}^{+} & + & {OH}^{-} \\ Initial conc. & 0.1 & 0 & 0 \\ At equilibrium & 0.1 (1 - α) & 0.1 α & 0.1 α \end{array}$
$∴ K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{4} OH]}$
$1.8 \times 10^{- 5} = \frac{(0.1 α) (0.1 α)}{0.1 (1 - α)}$
$1.8 \times 10^{- 5} = (0.1) α^{2} ......(i) [∵ α <<< 1]$
Now,
${[{OH}^{-}]}^{2} = (0.1)^{2} α^{2} = 0.1 [(0.1) α^{2}] \dots . . . (ii)$
From eqn (i) and (ii), we get
${[{OH}^{-}]}^{2} = 0.1 \times 1.8 \times 10^{- 5} = 1.8 \times 10^{- 6} M$

AMU-2015

Chemical Equilibrium

229258 Consider the following statements with regard to chemical equilibrium -
1.Equilibrium constant is dependent of initial concentrations of reactants and products.
2.Equilibrium constant is independent of initial concentrations of reactants and products.
3.Equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.
Which of the above statements are correct?

1 1 and 2 only

2 2 and 3 only

3 1 and 3 only

4 1,2 and 3

Explanation:

Application of equilibrium constant is given as -
(i\&ii) The value of equilibrium constant is independent of initial concentration of the reactant and product. (iii) Let us write equilibrium constant for the reaction,
$\begin{aligned} H_{2} (g) + I_{2} (g) ⇌ 2 HI (g) \\ ∴ & K_{c} = \frac{[HI]^{2}}{[H_{2}] [I_{2}]} \end{aligned}$
The equilibrium constant for the reverse reaction, $2 HI (g) ⇌ H_{2} (g) + I_{2} (g)$ , at the same temperature is,
$K_{c}^{'} = \frac{[H_{2}] [I_{2}]}{[HI]^{2}}$
From equation (i) and (ii)-
$K_{c}^{'} = \frac{1}{K_{c}}$
Hence, statement is true.

SCRA-2015

Chemical Equilibrium

229259 In a closed cylinder of capacity $24.6 L$ , the following reaction occurs at $27^{\circ} C$ .
$A_{2} (s) ⇌ B_{2} (s) + 2 C (g)$
At equilibrium, $1 g$ of $B_{2} (s)$ (molor mass $= 50 g$ mol $^{- 1}$ ) is present. The equilibrium constant $K_{p}$ for the equilibrium in ${atm}^{2}$ unit is
$(R = 0.082 L atm K^{- 1} {mol}^{- 1})$

1

1.6 \times 10^{- 2}

2

1.6 \times 10^{- 5}

3

1.6 \times 10^{- 3}

4

1.6 \times 10^{- 4}

5

1.6 \times 10^{- 1}

Explanation:

Given data : $V = 24.6 L, T = 27^{\circ} C$ amount of For the reaction,
$B_{2} = 1 g, K_{p} = ?$
$\begin{array}{ccc} A_{2} (s) & ⇌ & B_{2} (s) & + & 2 C (g) \\ 0 & 0 \\ (a - x) & x & 2 x \end{array}$
since, at equilibrium, $[B_{2}] = \frac{mass}{molor mass \times volume}$
$\begin{aligned} = \frac{1}{50 \times 24.6} \\ = & \frac{1}{1230} = 0.000813 = 8.13 \times 10^{- 4} \end{aligned}$
$K_{c} = [C]^{2} = {(2 \times 8.13 \times 10^{- 4})}^{2}$
$= (16.26)^{2} \times 10^{- 8}$
$= 264.38 \times 10^{- 8}$
$= 2.64 \times 10^{- 6}$
$K_{p} = K_{c} (RT)^{Δ n_{g}}$
$K_{p} = 2.64 \times 10^{- 6} (0.082 \times 300)^{2 - 0}$
$= 2.64 \times 10^{- 6} (24.6)^{2}$
$= 2.64 \times 605.16 \times 10^{- 6}$
$= 1597.62 \times 10^{- 6}$
$= 1.59 \times 10^{- 3}$
$\approx 1.6 \times 10^{- 3}$

Kerala-CEE-2014

Chemical Equilibrium

229262 Using the Gibbs energy change, $Δ G^{0} = + 63.3 kJ$ for the following reaction,
${Ag}_{2} {CO}_{3} (s) ⇌ {2 A g}^{+} (aq) + {CO}_{3}^{2 -} (aq)$
the $K_{sp}$ of ${Ag}_{2} {CO}_{3} (s)$ is water at $25^{\circ} C$ is $(R = 8.314 J K^{- 1} {mol}^{- 1}$ )

1

3.2 \times 10^{- 26}

2

8.0 \times 10^{- 12}

3

2.9 \times 10^{- 3}

4

7.9 \times 10^{- 2}

Explanation:

$Δ G^{o}$ is related to $K_{sp}$ by the equation, $Δ G^{o} = - 2.303 RT \log K_{sp}$ Given,
$\begin{aligned} Δ G^{o} & = + 63.3 kJ \\ = 63.3 \times 10^{3} J \end{aligned}$
Thus, Substitute $Δ G^{o} = 63.3 \times 10^{3} J$
$R = 8.314 J K^{- 1} {mol}^{- 1}$ and $T = 298 K$
From the above equation we get : -
$63.3 \times 10^{3} = - 2.303 \times 8.314 \times 298 \times \log K_{sp}$
$∴ \log K_{sp} = - 11.09$
$K_{sp} = antilog (- 11.09)$
$K_{sp} = 8.0 \times 10^{- 12}$

NEET-2014

Chemical Equilibrium

229257 What will be the value of ${[{OH}^{-}]}^{2}$ in the $0.1 M$ solution of ammonium hydroxide having $K_{b} =$ $1.8 \times 10^{- 5}$ ?

1

1.8 \times 10^{- 7}

2

1.8 \times 10^{- 6}

3

1.8 \times 10^{- 4}

4

1.8 \times 10^{- 3}

Explanation:

The dissociation reaction of ${NH}_{4} OH$ is -
$\begin{array}{cccc} {NH}_{4} OH & ⇌ & {NH}_{4}^{+} & + & {OH}^{-} \\ Initial conc. & 0.1 & 0 & 0 \\ At equilibrium & 0.1 (1 - α) & 0.1 α & 0.1 α \end{array}$
$∴ K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{4} OH]}$
$1.8 \times 10^{- 5} = \frac{(0.1 α) (0.1 α)}{0.1 (1 - α)}$
$1.8 \times 10^{- 5} = (0.1) α^{2} ......(i) [∵ α <<< 1]$
Now,
${[{OH}^{-}]}^{2} = (0.1)^{2} α^{2} = 0.1 [(0.1) α^{2}] \dots . . . (ii)$
From eqn (i) and (ii), we get
${[{OH}^{-}]}^{2} = 0.1 \times 1.8 \times 10^{- 5} = 1.8 \times 10^{- 6} M$

AMU-2015

Chemical Equilibrium

229258 Consider the following statements with regard to chemical equilibrium -
1.Equilibrium constant is dependent of initial concentrations of reactants and products.
2.Equilibrium constant is independent of initial concentrations of reactants and products.
3.Equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.
Which of the above statements are correct?

1 1 and 2 only

2 2 and 3 only

3 1 and 3 only

4 1,2 and 3

Explanation:

Application of equilibrium constant is given as -
(i\&ii) The value of equilibrium constant is independent of initial concentration of the reactant and product. (iii) Let us write equilibrium constant for the reaction,
$\begin{aligned} H_{2} (g) + I_{2} (g) ⇌ 2 HI (g) \\ ∴ & K_{c} = \frac{[HI]^{2}}{[H_{2}] [I_{2}]} \end{aligned}$
The equilibrium constant for the reverse reaction, $2 HI (g) ⇌ H_{2} (g) + I_{2} (g)$ , at the same temperature is,
$K_{c}^{'} = \frac{[H_{2}] [I_{2}]}{[HI]^{2}}$
From equation (i) and (ii)-
$K_{c}^{'} = \frac{1}{K_{c}}$
Hence, statement is true.

SCRA-2015

Chemical Equilibrium

229259 In a closed cylinder of capacity $24.6 L$ , the following reaction occurs at $27^{\circ} C$ .
$A_{2} (s) ⇌ B_{2} (s) + 2 C (g)$
At equilibrium, $1 g$ of $B_{2} (s)$ (molor mass $= 50 g$ mol $^{- 1}$ ) is present. The equilibrium constant $K_{p}$ for the equilibrium in ${atm}^{2}$ unit is
$(R = 0.082 L atm K^{- 1} {mol}^{- 1})$

1

1.6 \times 10^{- 2}

2

1.6 \times 10^{- 5}

3

1.6 \times 10^{- 3}

4

1.6 \times 10^{- 4}

5

1.6 \times 10^{- 1}

Explanation:

Given data : $V = 24.6 L, T = 27^{\circ} C$ amount of For the reaction,
$B_{2} = 1 g, K_{p} = ?$
$\begin{array}{ccc} A_{2} (s) & ⇌ & B_{2} (s) & + & 2 C (g) \\ 0 & 0 \\ (a - x) & x & 2 x \end{array}$
since, at equilibrium, $[B_{2}] = \frac{mass}{molor mass \times volume}$
$\begin{aligned} = \frac{1}{50 \times 24.6} \\ = & \frac{1}{1230} = 0.000813 = 8.13 \times 10^{- 4} \end{aligned}$
$K_{c} = [C]^{2} = {(2 \times 8.13 \times 10^{- 4})}^{2}$
$= (16.26)^{2} \times 10^{- 8}$
$= 264.38 \times 10^{- 8}$
$= 2.64 \times 10^{- 6}$
$K_{p} = K_{c} (RT)^{Δ n_{g}}$
$K_{p} = 2.64 \times 10^{- 6} (0.082 \times 300)^{2 - 0}$
$= 2.64 \times 10^{- 6} (24.6)^{2}$
$= 2.64 \times 605.16 \times 10^{- 6}$
$= 1597.62 \times 10^{- 6}$
$= 1.59 \times 10^{- 3}$
$\approx 1.6 \times 10^{- 3}$

Kerala-CEE-2014

Chemical Equilibrium

229262 Using the Gibbs energy change, $Δ G^{0} = + 63.3 kJ$ for the following reaction,
${Ag}_{2} {CO}_{3} (s) ⇌ {2 A g}^{+} (aq) + {CO}_{3}^{2 -} (aq)$
the $K_{sp}$ of ${Ag}_{2} {CO}_{3} (s)$ is water at $25^{\circ} C$ is $(R = 8.314 J K^{- 1} {mol}^{- 1}$ )

1

3.2 \times 10^{- 26}

2

8.0 \times 10^{- 12}

3

2.9 \times 10^{- 3}

4

7.9 \times 10^{- 2}

Explanation:

$Δ G^{o}$ is related to $K_{sp}$ by the equation, $Δ G^{o} = - 2.303 RT \log K_{sp}$ Given,
$\begin{aligned} Δ G^{o} & = + 63.3 kJ \\ = 63.3 \times 10^{3} J \end{aligned}$
Thus, Substitute $Δ G^{o} = 63.3 \times 10^{3} J$
$R = 8.314 J K^{- 1} {mol}^{- 1}$ and $T = 298 K$
From the above equation we get : -
$63.3 \times 10^{3} = - 2.303 \times 8.314 \times 298 \times \log K_{sp}$
$∴ \log K_{sp} = - 11.09$
$K_{sp} = antilog (- 11.09)$
$K_{sp} = 8.0 \times 10^{- 12}$

NEET-2014

Chemical Equilibrium

229257 What will be the value of ${[{OH}^{-}]}^{2}$ in the $0.1 M$ solution of ammonium hydroxide having $K_{b} =$ $1.8 \times 10^{- 5}$ ?

1

1.8 \times 10^{- 7}

2

1.8 \times 10^{- 6}

3

1.8 \times 10^{- 4}

4

1.8 \times 10^{- 3}

Explanation:

The dissociation reaction of ${NH}_{4} OH$ is -
$\begin{array}{cccc} {NH}_{4} OH & ⇌ & {NH}_{4}^{+} & + & {OH}^{-} \\ Initial conc. & 0.1 & 0 & 0 \\ At equilibrium & 0.1 (1 - α) & 0.1 α & 0.1 α \end{array}$
$∴ K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{4} OH]}$
$1.8 \times 10^{- 5} = \frac{(0.1 α) (0.1 α)}{0.1 (1 - α)}$
$1.8 \times 10^{- 5} = (0.1) α^{2} ......(i) [∵ α <<< 1]$
Now,
${[{OH}^{-}]}^{2} = (0.1)^{2} α^{2} = 0.1 [(0.1) α^{2}] \dots . . . (ii)$
From eqn (i) and (ii), we get
${[{OH}^{-}]}^{2} = 0.1 \times 1.8 \times 10^{- 5} = 1.8 \times 10^{- 6} M$

AMU-2015

Chemical Equilibrium

229258 Consider the following statements with regard to chemical equilibrium -
1.Equilibrium constant is dependent of initial concentrations of reactants and products.
2.Equilibrium constant is independent of initial concentrations of reactants and products.
3.Equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.
Which of the above statements are correct?

1 1 and 2 only

2 2 and 3 only

3 1 and 3 only

4 1,2 and 3

Explanation:

Application of equilibrium constant is given as -
(i\&ii) The value of equilibrium constant is independent of initial concentration of the reactant and product. (iii) Let us write equilibrium constant for the reaction,
$\begin{aligned} H_{2} (g) + I_{2} (g) ⇌ 2 HI (g) \\ ∴ & K_{c} = \frac{[HI]^{2}}{[H_{2}] [I_{2}]} \end{aligned}$
The equilibrium constant for the reverse reaction, $2 HI (g) ⇌ H_{2} (g) + I_{2} (g)$ , at the same temperature is,
$K_{c}^{'} = \frac{[H_{2}] [I_{2}]}{[HI]^{2}}$
From equation (i) and (ii)-
$K_{c}^{'} = \frac{1}{K_{c}}$
Hence, statement is true.

SCRA-2015

Chemical Equilibrium

229259 In a closed cylinder of capacity $24.6 L$ , the following reaction occurs at $27^{\circ} C$ .
$A_{2} (s) ⇌ B_{2} (s) + 2 C (g)$
At equilibrium, $1 g$ of $B_{2} (s)$ (molor mass $= 50 g$ mol $^{- 1}$ ) is present. The equilibrium constant $K_{p}$ for the equilibrium in ${atm}^{2}$ unit is
$(R = 0.082 L atm K^{- 1} {mol}^{- 1})$

1

1.6 \times 10^{- 2}

2

1.6 \times 10^{- 5}

3

1.6 \times 10^{- 3}

4

1.6 \times 10^{- 4}

5

1.6 \times 10^{- 1}

Explanation:

Given data : $V = 24.6 L, T = 27^{\circ} C$ amount of For the reaction,
$B_{2} = 1 g, K_{p} = ?$
$\begin{array}{ccc} A_{2} (s) & ⇌ & B_{2} (s) & + & 2 C (g) \\ 0 & 0 \\ (a - x) & x & 2 x \end{array}$
since, at equilibrium, $[B_{2}] = \frac{mass}{molor mass \times volume}$
$\begin{aligned} = \frac{1}{50 \times 24.6} \\ = & \frac{1}{1230} = 0.000813 = 8.13 \times 10^{- 4} \end{aligned}$
$K_{c} = [C]^{2} = {(2 \times 8.13 \times 10^{- 4})}^{2}$
$= (16.26)^{2} \times 10^{- 8}$
$= 264.38 \times 10^{- 8}$
$= 2.64 \times 10^{- 6}$
$K_{p} = K_{c} (RT)^{Δ n_{g}}$
$K_{p} = 2.64 \times 10^{- 6} (0.082 \times 300)^{2 - 0}$
$= 2.64 \times 10^{- 6} (24.6)^{2}$
$= 2.64 \times 605.16 \times 10^{- 6}$
$= 1597.62 \times 10^{- 6}$
$= 1.59 \times 10^{- 3}$
$\approx 1.6 \times 10^{- 3}$

Kerala-CEE-2014

Chemical Equilibrium

229262 Using the Gibbs energy change, $Δ G^{0} = + 63.3 kJ$ for the following reaction,
${Ag}_{2} {CO}_{3} (s) ⇌ {2 A g}^{+} (aq) + {CO}_{3}^{2 -} (aq)$
the $K_{sp}$ of ${Ag}_{2} {CO}_{3} (s)$ is water at $25^{\circ} C$ is $(R = 8.314 J K^{- 1} {mol}^{- 1}$ )

1

3.2 \times 10^{- 26}

2

8.0 \times 10^{- 12}

3

2.9 \times 10^{- 3}

4

7.9 \times 10^{- 2}

Explanation:

$Δ G^{o}$ is related to $K_{sp}$ by the equation, $Δ G^{o} = - 2.303 RT \log K_{sp}$ Given,
$\begin{aligned} Δ G^{o} & = + 63.3 kJ \\ = 63.3 \times 10^{3} J \end{aligned}$
Thus, Substitute $Δ G^{o} = 63.3 \times 10^{3} J$
$R = 8.314 J K^{- 1} {mol}^{- 1}$ and $T = 298 K$
From the above equation we get : -
$63.3 \times 10^{3} = - 2.303 \times 8.314 \times 298 \times \log K_{sp}$
$∴ \log K_{sp} = - 11.09$
$K_{sp} = antilog (- 11.09)$
$K_{sp} = 8.0 \times 10^{- 12}$

NEET-2014

Chemical Equilibrium

229257 What will be the value of ${[{OH}^{-}]}^{2}$ in the $0.1 M$ solution of ammonium hydroxide having $K_{b} =$ $1.8 \times 10^{- 5}$ ?

1

1.8 \times 10^{- 7}

2

1.8 \times 10^{- 6}

3

1.8 \times 10^{- 4}

4

1.8 \times 10^{- 3}

Explanation:

The dissociation reaction of ${NH}_{4} OH$ is -
$\begin{array}{cccc} {NH}_{4} OH & ⇌ & {NH}_{4}^{+} & + & {OH}^{-} \\ Initial conc. & 0.1 & 0 & 0 \\ At equilibrium & 0.1 (1 - α) & 0.1 α & 0.1 α \end{array}$
$∴ K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{4} OH]}$
$1.8 \times 10^{- 5} = \frac{(0.1 α) (0.1 α)}{0.1 (1 - α)}$
$1.8 \times 10^{- 5} = (0.1) α^{2} ......(i) [∵ α <<< 1]$
Now,
${[{OH}^{-}]}^{2} = (0.1)^{2} α^{2} = 0.1 [(0.1) α^{2}] \dots . . . (ii)$
From eqn (i) and (ii), we get
${[{OH}^{-}]}^{2} = 0.1 \times 1.8 \times 10^{- 5} = 1.8 \times 10^{- 6} M$

AMU-2015

Chemical Equilibrium

229258 Consider the following statements with regard to chemical equilibrium -
1.Equilibrium constant is dependent of initial concentrations of reactants and products.
2.Equilibrium constant is independent of initial concentrations of reactants and products.
3.Equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.
Which of the above statements are correct?

1 1 and 2 only

2 2 and 3 only

3 1 and 3 only

4 1,2 and 3

Explanation:

Application of equilibrium constant is given as -
(i\&ii) The value of equilibrium constant is independent of initial concentration of the reactant and product. (iii) Let us write equilibrium constant for the reaction,
$\begin{aligned} H_{2} (g) + I_{2} (g) ⇌ 2 HI (g) \\ ∴ & K_{c} = \frac{[HI]^{2}}{[H_{2}] [I_{2}]} \end{aligned}$
The equilibrium constant for the reverse reaction, $2 HI (g) ⇌ H_{2} (g) + I_{2} (g)$ , at the same temperature is,
$K_{c}^{'} = \frac{[H_{2}] [I_{2}]}{[HI]^{2}}$
From equation (i) and (ii)-
$K_{c}^{'} = \frac{1}{K_{c}}$
Hence, statement is true.

SCRA-2015

Chemical Equilibrium

229259 In a closed cylinder of capacity $24.6 L$ , the following reaction occurs at $27^{\circ} C$ .
$A_{2} (s) ⇌ B_{2} (s) + 2 C (g)$
At equilibrium, $1 g$ of $B_{2} (s)$ (molor mass $= 50 g$ mol $^{- 1}$ ) is present. The equilibrium constant $K_{p}$ for the equilibrium in ${atm}^{2}$ unit is
$(R = 0.082 L atm K^{- 1} {mol}^{- 1})$

1

1.6 \times 10^{- 2}

2

1.6 \times 10^{- 5}

3

1.6 \times 10^{- 3}

4

1.6 \times 10^{- 4}

5

1.6 \times 10^{- 1}

Explanation:

Given data : $V = 24.6 L, T = 27^{\circ} C$ amount of For the reaction,
$B_{2} = 1 g, K_{p} = ?$
$\begin{array}{ccc} A_{2} (s) & ⇌ & B_{2} (s) & + & 2 C (g) \\ 0 & 0 \\ (a - x) & x & 2 x \end{array}$
since, at equilibrium, $[B_{2}] = \frac{mass}{molor mass \times volume}$
$\begin{aligned} = \frac{1}{50 \times 24.6} \\ = & \frac{1}{1230} = 0.000813 = 8.13 \times 10^{- 4} \end{aligned}$
$K_{c} = [C]^{2} = {(2 \times 8.13 \times 10^{- 4})}^{2}$
$= (16.26)^{2} \times 10^{- 8}$
$= 264.38 \times 10^{- 8}$
$= 2.64 \times 10^{- 6}$
$K_{p} = K_{c} (RT)^{Δ n_{g}}$
$K_{p} = 2.64 \times 10^{- 6} (0.082 \times 300)^{2 - 0}$
$= 2.64 \times 10^{- 6} (24.6)^{2}$
$= 2.64 \times 605.16 \times 10^{- 6}$
$= 1597.62 \times 10^{- 6}$
$= 1.59 \times 10^{- 3}$
$\approx 1.6 \times 10^{- 3}$

Kerala-CEE-2014

Chemical Equilibrium

229262 Using the Gibbs energy change, $Δ G^{0} = + 63.3 kJ$ for the following reaction,
${Ag}_{2} {CO}_{3} (s) ⇌ {2 A g}^{+} (aq) + {CO}_{3}^{2 -} (aq)$
the $K_{sp}$ of ${Ag}_{2} {CO}_{3} (s)$ is water at $25^{\circ} C$ is $(R = 8.314 J K^{- 1} {mol}^{- 1}$ )

1

3.2 \times 10^{- 26}

2

8.0 \times 10^{- 12}

3

2.9 \times 10^{- 3}

4

7.9 \times 10^{- 2}

Explanation:

$Δ G^{o}$ is related to $K_{sp}$ by the equation, $Δ G^{o} = - 2.303 RT \log K_{sp}$ Given,
$\begin{aligned} Δ G^{o} & = + 63.3 kJ \\ = 63.3 \times 10^{3} J \end{aligned}$
Thus, Substitute $Δ G^{o} = 63.3 \times 10^{3} J$
$R = 8.314 J K^{- 1} {mol}^{- 1}$ and $T = 298 K$
From the above equation we get : -
$63.3 \times 10^{3} = - 2.303 \times 8.314 \times 298 \times \log K_{sp}$
$∴ \log K_{sp} = - 11.09$
$K_{sp} = antilog (- 11.09)$
$K_{sp} = 8.0 \times 10^{- 12}$

NEET-2014

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