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_(2)O(g) rarr Cu_(2)O(s) + H_(2)(g)` is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln is ____. (Given: total pressure = 1 bar, R (universal gas constant) = `8 J K−1 mol^(−1), ln(10) = 2.3. Cu(s) and Cu_(2)O(s)` are mutually immiscible. At 1250 `K: 2Cu(s) + 1//2 O_(2)(g) rarr Cu_(2)O(s)`
`triangle H^(theta) = − 78,000 J mol^(−1)`
`H_(2)(g) + 1//2 O_(2)(g) rarr H_(2)O(g), triangle G^(theta) = − 1,78,000 J mol^(−1)`, G is the Gibbs energy

Answer

A, D

Explanation

`2Cu(s)+(1)/(4)O_(2)(g)rarr1Cu_(2)O(s)" "DeltaG^(@)=-78 kJ`
`[H_(2)(g)+(1)/(2)O_(2)rarrH_(2)O(g)" "DeltaG^(@)=-178 kJ]xx(-1)`
Hence, `2Cu(s)+H_(2)O(g)rarrCu_(2)O+H_(2)(g)DeltaG^(@)=+100 kJ`
`DeltaG=DeltaG^(@)+RT ln Q`
`0=+100+(8)/(1000)xx1250ln.(P_(H_(2)))/(P_(H_(2)O))`
`-(100xx1000)/(8)=1250ln.P_(H_(2))/(((1)/(100)xx1))`
`ln P_(H_(2))=-14.6`