At 298 K, the equilibrium constant is 2×10^15 for the reaction: Cus+2Ag^(+)aq⇌Cu^(2+)aq+2Ags The equilibrium constant for the reaction 1/2Cu^(2+)aq+Ags⇌1/2Cus+Ag^(+)aq is x×10^(-8). The value of x is ______. (Round off the answer to the nearest integer)

Q26 / 90 · JEE Main 2022 (26 Jul Shift 1) · Chemistry

At $298 K$ , the equilibrium constant is $2 \times 10^{15}$ for the reaction:
$Cu (s) + 2 {Ag}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + 2 Ag (s)$
The equilibrium constant for the reaction
$\frac{1}{2} {Cu}^{2 +} (aq) + Ag (s) ⇌ \frac{1}{2} Cu (s) + {Ag}^{+} (aq)$ is $x \times 10^{- 8}$ . The value of $x$ is ______. (Round off the answer to the nearest integer)

Correct answer: 2

Solution

$Cu (s) + 2 {Ag}^{+} (aq) ⇌ {Cu}^{2 +} (aq) + 2 Ag (s); K_{eq} = 2 \times 10^{15}$
$\frac{1}{2} {Cu}^{2 +} (aq) + Ag (s) ⇌ \frac{1}{2} Cu (s) + {Ag}^{+} (aq)$
The equilibrium reaction is reversed and multiplied with $\frac{1}{2}$ ,
Hence, $K_{eq}^{'} = \frac{1}{\sqrt{K_{eq}}} = \frac{1}{\sqrt{2 \times 10^{15}}} = x \times 10^{- 8}$
$\Rightarrow \frac{1}{\sqrt{20}} \times \frac{1}{10^{7}} = x \times 10^{- 8}$
$\Rightarrow \frac{1}{\sqrt{20}} \times 10^{- 7} = x \times 10^{- 8}$
$\frac{10}{\sqrt{20}} = x$
$\Rightarrow x = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} = 2.236$
$≃ 2.24$

Hence, the correct answer is 2.

At 298 K, the equilibrium constant is 2×1015 for the reaction:Cus+2Ag+aq⇌Cu2+aq+2AgsThe equilibrium constant for the reaction12Cu2+aq+Ags⇌12Cus+Ag+aq is x×10-8. The value of x is ______. (Round off the answer to the nearest integer)

Solution

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