JEE Advanced 2018 — Paper 2 — Question 24

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 $1 \mathrm{~mole} \%$ of water vapour as

impurity. The water vapour oxidises copper as per the reaction given below:

$2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g})$

$\mathrm{p}_{\mathrm{H}_{2}}$ is the minimum partial pressure

of $\mathrm{H}_{2}$ (in bar) needed to prevent the oxidation at 1250 K . The value of $\ln \left(\mathrm{p}_{\mathrm{H}_{2}}\right)$ 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.

$$\text { At } 1250 \mathrm{~K}: 2 \mathrm{Cu}(\mathrm{~s})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{~s}) ; \Delta \mathrm{G}^{\theta}=-78,000 \mathrm{~J} \mathrm{~mol}^{-1}$$ $$\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) ; \Delta \mathrm{G}^{\theta}=-1,78,000 \mathrm{~J} \mathrm{~mol}^{-1} ; G \text { is the Gibbs energy) }$$