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:

$2Cu\left(s\right)+ \, H_{2}O\left(g\right) \rightarrow \, Cu_{2}O\left(\right.s\left.\right) \, + \, H_{2}\left(\right.g\left.\right)$

$P_{H_{2}}$ 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 \left(10\right) \, = \, 2.3. \, Cu\left(\right.s\left.\right) \, and \, Cu_{2}O\left(\right.s\left.\right) \, $ are mutually immiscible.

At $1250 \, K:2Cu\left(s\right)+\frac{1}{2} \, O_{2}\left(g\right) \rightarrow \, Cu_{2}O\left(\right.s\left.\right); \, \Delta G^{\ominus} \, = \, - \, 78,000 \, J \, mol^{- 1}$

$H_{2}\left(g\right)+\frac{1}{2}O_{2}\left(g\right) \rightarrow \, H_{2}O\left(\right.g\left.\right); \, \, \left(\Delta \right)^{\ominus} \, = \, - \, 1,78,000 \, J \, mol^{- 1}; \, G \, $ is the Gibbs energy)

Answer 1

$2Cu\left(s\right)+\frac{1}{4}O_{2}\left(g\right) \rightarrow 1 \, Cu_{2}O\left(s\right)$ $\Delta G^{0}=-78 \, kJ$

$\left[H_{2} \left(g\right) + \frac{1}{2} O_{2} \rightarrow H_{2} O \left(g\right) \, \, \Delta G^{o} = - 178 \, k J\right]\times \left(- 1\right)$

Hence, $2Cu\left(s\right)+H_{2}O\left(g\right) \rightarrow Cu_{2}O+H_{2}\left(g\right) \, \, \, \Delta G^{o}=+100kJ$

$\Delta G=\Delta G^{o}+RTln Q$

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

$-\frac{100 \times 1000}{8}=1250ln \frac{\left(P_{H}\right)_{2}}{\left(\frac{1}{100} \times 1\right)}$

$ln P_{H_{2}} = - 14.6$