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

`-14,6`

Explanation

`2Cu = 1/2 O_(2)(g) rarr Cu_(2)O(s), Delta G_(1)^(@) = -78 kJ" "….(i)`
`H_(2)(g) + 1/2O_(2)(g) rarr H_(2)O(g), DeltaG_(2)^(@) = -178 kJ" "…..(ii)`
`2Cu(s) + H_(2)O (g) rarr Cu_(2) O(s) + H_(2)(g), Delta G_(3)^(@) = 100 kJ" ".....(iii)`
Eq.(i) Eq.(ii) = Eq.(iii)
`implies 0 = 100 + 8/1000 xx 1250 ln (P_(H_2))/(P_(H_2O)) implies -(10^5)/8 = 1250 ln (P_(H_2))/((1)/(100) xx 1) (1 % "of" 1 "bar")`
`implies -10 = ln P_(H_2) xx 100 implies -10 = ln P_(H_2) xx ln 10^(2)`
`P_(H_2) = - 10 - 4.6 = -14.6`.