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) 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 |
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Answer» Given, `2Cu(s)+(1)/(2)O_(2)(g)toCu_(2)O,DeltaG^(@)=-78kJ" "mol^(-1)` `H_(2)O(g)toH_(2)(g)+(1)/(2)O_(2)(g)," "DeltaG^(@)=+178kJ" "mol^(-1)` `overline(2Cu(s)+H_(2)(g)hArrCu_(2)O(s)+H_(2)(g),DeltaG^(@)=+100kJ" "mol^(-1)` `overline(p_(H_(2)O)=(1)/(100)xx1=0.01`bar we know, `DeltaG=DeltaG^(@)+RT" ln "Q` . . .(i) At equilibrium `DeltaG=0` `Q-K=(p_(H_(2)))/(p_(H_(2)O))=(pH_(2))/(0.01)` `R=8JK^(-1)mol^(-1)` `T=1250K` Putting values in (i) `0=+100kJ+(8)/(1000)xx1250" log"_(e)(p_(H_(2)))/(0.01)` `log_(e)pH_(2)=-14.6` |
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