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:
$2Cu(s) + H_2O(g) \to Cu_2O(s) + H_2(g)$
$p_{H_2}$ 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 $Cu_2O(s)$ are mutually immiscible.
At $1250 \, K: 2Cu(s) + ½ O_2(g) \to Cu_2O(s); ΔG^Ɵ = − 78,000 \, J\, mol^{−1}$
$H_2(g) + ½ O_2(g) \to H_2O(g); ΔG^Ɵ = − 1,78,000 \, J \, mol^{−1};$ G is the Gibbs energy)

Answer 1

$2 Cu ( s )+\frac{1}{4} O _{2}( g ) \rightarrow 1 Cu _{2} O ( s ) $
$ \Delta G ^{0}=-78 \,kJ $
${\left[ H _{2}( g )+\frac{1}{2} O _{2} \rightarrow H _{2} O ( g )\right.} $
$ \left.\Delta G^{0}=-178 \,kJ \right] \times(-1)$
Hence, $2 Cu ( s )+ H _{2} O ( g ) \rightarrow Cu _{2} O + H _{2}( g ) $
$\Delta G ^{0}=+100 \,kJ$
$0=+100+\frac{8}{1000} \times 1250 \ln \frac{P_{H_{2}}}{P_{H_{2} O }}$
$-\frac{100 \times 1000}{8}=1250 \ln \frac{P_{H_{2}}}{\left(\frac{1}{100} \times 1\right)}$
$\ln P_{H}=-14.6$