Question

At $298\ K$, the equilibrium constant is $2 × 10^{15}$ for the reaction:
$Cu(s)+2Ag+(Aq)⇌Cu^{2+}(aq)+2Ag(s)$
The equilibrium constant for the reaction:
$\frac 12Cu^{2+}(aq)+Ag(s)+(Aq)⇌\frac 12Cu(s)+Ag^+(aq)$ is $x \times10^{-8}$.
The value of $x$ is ______ . (Nearest integer)

Answer 1

$Cu(s)+2Ag^+(Aq)⇌Cu^{2+}(aq)+2Ag(s)$
Equilibrium constant, $k = 2 × 10^{15}$
$\frac 12Cu(s)+Ag+(aq)⇌Cu^{2+}(aq)+2Ag(s)$
$k'=\frac {1}{(k)^{\frac 12}}$

$k'=\frac {1}{(2×10^{15})^{\frac 12}}$

$k'=2.23×10^{–8}$
$k≃2$

So, the answer is $2$.