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)rarrCu_(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 1250 K. The value of ln(P_(H_(2))) is_____ (Given : total pressure = 1 bar, R (universal gas constant)= 8 JK^(-1) mol^(-1), ln(10) = 2.3, Cu(s) and Cu_(2)O(s) are mutually immiscible.At 1250K :2Cu(s)+1//2O_(2)(g)rarrCu_(2)O(s),DeltaG^(theta)=-78,000J mol^(-1)H_(2)(g)+1//2O_(2)(g)rarrH_(2)O(g),DeltaG^(theta)=-1,78,000 J mol^(-1), G the Gibbs energy

The surface of copper gets tarnished by the formation of copper oxide.

1.	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)rarrCu_(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 1250 K. The value of ln(P_(H_(2))) is_____ (Given : total pressure = 1 bar, R (universal gas constant)= 8 JK^(-1) mol^(-1), ln(10) = 2.3, Cu(s) and Cu_(2)O(s) are mutually immiscible.At 1250K :2Cu(s)+1//2O_(2)(g)rarrCu_(2)O(s),DeltaG^(theta)=-78,000J mol^(-1)H_(2)(g)+1//2O_(2)(g)rarrH_(2)O(g),DeltaG^(theta)=-1,78,000 J mol^(-1), G the Gibbs energy
Answer» Solution :`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`