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 per cent of water vapour as impurity. The water vapour oxidises copper as par the reaction given below:
`2Cu(s)+H_(2)O (g) to Cu_(2)O (s)+H_(2)(g)`
`p_(H_(2))` is the minimum partical pressure of `H_2` (1 bar) needed to prevent the oxidation at 1250 K. The value of `In(P_(H_(2))` is... [Given: Total pressure =1 bar]
R (universal gas constant) `=8 JK^(-1)" mol"^(-1), In (10) "=2.3, Cu(s) and "Cu_2O(s)` are mutually immiscible.
At `1250 K : 2Cu(s) +1/2 O_(2) (g) to Cu_(2) O(s), triangleG^(@)=-78000J mol^(-1)`
`H_(2) (g)+1/2 O_(2) (g) to H_(2)O , triangleG^@=-1,78,000 J mol^(-1)`

Answer

`-14.6`

Explanation

From the given equation,
`2Cu(s)+H_(2)O (g) to Cu_(2)O(s)+H_(2)(g), triangleG^@=[phone]s=100000`
Now, `triangleG^@=-RT ln (p_(H_2))/(p_(H_2)O)`
`100000=-8 xx 1250 (ln p_(H_2))-ln 0.01)`
Calculate ln `p_(H_(2)) (ln 0.01=2.3 log 0.01)]`