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 $1250K$. However, the $N_2$ gas contains $1$ mole $\%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
$2Cu(s)+H_{2}O(g)\to C{u}_{2}O(s)+H_2(g)$
$P_{H_2}$ Is the minimum partial pressure of $H_2$ (in bar) needed to prevent the oxidation at $1250K$. The value of $\ln \left(P_{H_2}\right)$ is
(Given: total pressure $=1$ bar, R (universal gas constant) $=8J{K}^{-1}mo{l}^{-1},\ln (10)=\mathrm{2.3.}Cu(s)$ and $C{u}_{2}O(s)$ are mutually immiscible.
At $1250K:2Cu(s)+\frac{1}{2}{O}_2(g)\to C{u}_{2}O(s);\Delta G=$ $H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(g);\Delta G=-1,78,000Jmo{l}^{-1};)$
Round off the answer up to the nearest integer.

Answer 1

Answer: 15

$2Cu\left(s\right)+\frac{1}{4}{O}_2\left(g\right)\to 1{\left(Cu\right)}_{2}O\left(s\right)$ $\Delta G^0=-78kJ$
$\left[H_2\left(g\right)+\frac{1}{2}{O}_2\to H_{2}O\left(g\right)\Delta G^o=-178kJ\right]\times \left(-1\right)$
Hence, $2Cu\left(s\right)+H_{2}O\left(g\right)\to {\left(Cu\right)}_{2}O+H_2\left(g\right)\Delta G^o=+100kJ$
$\Delta G=\Delta G^o+RTlnQ$
$0=+100+\frac{8}{1000}\times 1250ln\frac{P_{H_2}}{P{H}_{2}O}$
$-\frac{100\times 1000}{8}=1250\ln \frac{P_{H_2}}{\left(\frac{1}{100}\times 1\right)}$
$ln{P}_{H_2}=-14.6$
If the above value is rounded off to two significant figures it becomes $-15$.