Question

The surface of copper gets tarnished by the formation of copper oxide. $N _{2}$ gas was passed to prevent the oxide formation during heating of copper at $1250 \,K$. However, the $N _{2}$ gas contains $1$ mole $\%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
$2 Cu ( s )+ H _{2} O ( g ) \rightarrow Cu _{2} O ( s )+ H _{2}( g )$
$P _{ H _{2}}$ Is the minimum partial pressure of $H _{2}$ (in bar) needed to prevent the oxidation at $1250 \, K$. The value of $\ln \left( P _{ H _{2}}\right)$ is
(Given: total pressure $=1$ bar, R (universal gas constant) $=8 \,J \,K ^{-1} \,mol ^{-1}, \ln (10)=2.3 . Cu ( s )$ and $Cu _{2} O ( s )$ are mutually immiscible.
At $1250 \, K : 2 Cu ( s )+\frac{1}{2} O _{2}( g ) \rightarrow Cu _{2} O ( s ) ; \Delta G =$ $\left.H _{2}( g )+\frac{1}{2} O _{2}( g ) \rightarrow H _{2} O ( g ) ; \Delta G =-1,78,000 \, J \,mol ^{-1} ;\right)$
Round off the answer up to the nearest integer.

Answer 1

$2Cu\left(s\right)+\frac{1}{4}O_{2}\left(g\right) \rightarrow 1 \, \left(Cu\right)_{2}O\left(s\right)$ $\Delta G^{0}=-78 \, kJ$
$\left[H_{2} \left(g\right) + \frac{1}{2} O_{2} \rightarrow H_{2} O \left(g\right) \Delta G^{o} = - 178 \, kJ\right]\times \left(- 1\right)$
Hence, $2Cu\left(s\right)+H_{2}O\left(g\right) \rightarrow \left(Cu\right)_{2}O+H_{2}\left(g\right)\Delta G^{o}=+100\,kJ$
$\Delta G=\Delta G^{o}+RTln Q$
$0=+100+\frac{8}{1000}\times 1250ln \frac{P_{H_{2}}}{PH_{2} O}$
$-\frac{100 \times 1000}{8}=1250 \ln \frac{ P _{ H _{2}}}{\left(\frac{1}{100} \times 1\right)}$
$ln P_{H_{2}}=-14.6$
If the above value is rounded off to two significant figures it becomes $-15$.