JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Redox Reactions and Electrochemistry

The surface of copper gets tarnished by the formation of copper oxide. ${\mathrm{N}}_2$ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the ${\mathrm{N}}_2$ gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:

$2\mathrm{Cu}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\to {\mathrm{Cu}}_2\mathrm{O}(\mathrm{s})+{\mathrm{H}}_2(\mathrm{g})$

$p_{{\mathrm{H}}_2}$ is the minimum partial pressure of ${\mathrm{H}}_2$ (in bar) needed to prevent the oxidation at 1250 K. The value of $\ln (p_{{\mathrm{H}}_2})$ is ____.

(Given: total pressure = 1 bar, $R$ (universal gas constant) = $8\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$, $\ln (10)=2.3$. $\mathrm{Cu}(\mathrm{s})$ and ${\mathrm{Cu}}_2\mathrm{O}(\mathrm{s})$ are mutually immiscible.

At 1250 K:
$2\mathrm{Cu}(\mathrm{s})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{Cu}}_2\mathrm{O}(\mathrm{s})$; $\Delta G^{\theta }=-78,000\mathrm{J}{\mathrm{mol}}^{-1}$

${\mathrm{H}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{H}}_2\mathrm{O}(\mathrm{g})$; $\Delta G^{\ominus }=-1,78,000\mathrm{J}{\mathrm{mol}}^{-1}$; $G$ is the Gibbs energy)