Question

The surface of copper gets tarnished by the formation of copper oxide. $N_2$ gets was passed to prevent the oxide formation during heating of copper at $1250K$. However, the $N_2$ gas contains $1$ mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
$2Cu\left(s\right)+H_{2}O\left(g\right)\to C{u}_{2}O\left(s\right)+H_2\left(g\right)$
${\rho }_{H_2}$ is the minimum partial pressure of $H_2$ (in bar) needed to prevent the oxidation at $1250K$. The value of $ln\left({\rho }_{H_2}\right)$ is _________.

Given:

Total pressure $=1bar$

$R$ (universal gas constant) $=8J{K}^{-1}mo{l}^{-1}$

$ln\left(10\right)=2.3$

$Cu\left(s\right)$ and $C{u}_{2}O\left(s\right)$ are mutually immiscible
At $1250K$:

$2Cu\left(s\right)+1/2O_2\left(g\right)\to C{u}_{2}O\left(s\right);\Delta G^{\circ }=-78,000Jmo{l}^{-1}$ $H_2\left(g\right)+1/2O_2\left(g\right)\to H_{2}O\left(g\right);\Delta G^{\circ }=-1,78,000Jmo{l}^{-1}$

[Upto one decimal point and take modulus of the answer]

Answer 1

$2Cu\left(s\right)+\frac{1}{4}{O}_2\left(g\right)\to C{u}_{2}O\left(s\right)\Delta G^{\circ }=-78kJ$
$\left[H_2\right(g)+\frac{1}{2}{O}_2\to H_{2}O(g)\Delta G^{\circ }=-178kJ]\times (-1)$
Hence, $2Cu\left(s\right)+H_{2}O\left(g\right)\to C{u}_{2}O+H_2\left(g\right)\Delta G^{\circ }=+100kJ$

$\Delta G=\Delta G^{\circ }+RTlnQ$

$0=+100+\frac{8}{1000}\times 1250ln\frac{{\rho }_{H_2}}{{\rho }_{H_{2}O}}$

$-\frac{100\times 1000}{8}=1250ln\frac{{\rho }_{H_2}}{(\frac{1}{100}\times 1)}$

$\therefore ln{\rho }_{H_2}=-14.6$.