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:

$2Cu\left(s\right)+H_{2}O\left(g\right)\to C{u}_{2}O(s)+H_2(g)$

$P_{H_2}$ 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) $=8J{K}^{-1}mo{l}^{-1},ln\left(10\right)=\mathrm{2.3.}Cu(s)andC{u}_{2}O(s)$ are mutually immiscible.

At $1250K:2Cu\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to C{u}_{2}O(s);\Delta G^{\ominus }=-78,000Jmo{l}^{-1}$

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O(g);{\left(\Delta \right)}^{\ominus }=-1,78,000Jmo{l}^{-1};G$ is the Gibbs energy)

Answer 1

Answer: -14.6

$2Cu\left(s\right)+\frac{1}{4}{O}_2\left(g\right)\to 1C{u}_{2}O\left(s\right)$ $\Delta G^0=-78kJ$

$\left[H_2\left(g\right)+\frac{1}{2}{O}_2\to H_{2}O\left(g\right)\Delta G^o=-178kJ\right]\times \left(-1\right)$

Hence, $2Cu\left(s\right)+H_{2}O\left(g\right)\to C{u}_{2}O+H_2\left(g\right)\Delta G^o=+100kJ$

$\Delta G=\Delta G^o+RTlnQ$

$0=+100+\frac{8}{1000}\times 1250ln\frac{P_{H_2}}{P{H}_{2}O}$

$-\frac{100\times 1000}{8}=1250ln\frac{{\left(P_H\right)}_2}{\left(\frac{1}{100}\times 1\right)}$

$ln{P}_{H_2}=-14.6$