InterviewSolution
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InterviewSolution
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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 |
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Answer» Correct Answer - A::D `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` |
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