At 298 K, the equilibrium constant is $2 \times 10^{15}$ for the reaction: $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ The equilibrium constant for the reaction $\frac{1}{2} \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \rightleftharpoons \frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq})$ is $x \times 10^{-8}$. The value of x is ______. (Nearest Integer)

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Redox Reactions and Electrochemistry

At 298 K, the equilibrium constant is

2 \times 1 0^{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)

x \times 1 0^{- 8}

. The value of x is ______.
(Nearest Integer)

Answer: 2

The second reaction is the reverse of the first, divided by 2. When a reaction is reversed,

K^{'} = \frac{1}{K}

. When divided by a factor,

K^{''} = (K^{'})^{1/ n}

.

Here:

K^{'} = \frac{1}{K} = \frac{1}{2 \times 1 0 ^{15}}

K^{'} = \frac{1}{2 \times 1 0 ^{7.5}} = \frac{1}{2 \times 10 \times 1 0 ^{7}} = \frac{1}{20} \times 1 0^{- 7}

K^{'} = \frac{10}{20} \times 1 0^{- 8} = \frac{100}{20} \times 1 0^{- 8} = 5 \times 1 0^{- 8} \approx 2.24 \times 1 0^{- 8}

.

Nearest integer: 2.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer