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Thermodynamics question

2018 · Shift 2 · Q8
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Thermodynamics question

2018 · Shift 2 · Q8

JEE AdvancedChemistryThermodynamicsNumerical+3 / −1
The surface of copper gets tarnished by the formation of copper oxide. N2{N_2}N2​ gas was passed to prevent the oxide formation during heating of copper at 1250K.1250K.1250K. However, the N2{N_2}N2​ gas contains 111 mole % of water vapor as impurity. The water vapor oxidises copper as per the reaction given below : 2Cu(s)+H2O(g)→Cu2O(s)+H2(g)2Cu\left( s \right) + {H_2}O\left( g \right) \to C{u_2}O\left( s \right) + {H_2}\left( g \right)2Cu(s)+H2​O(g)→Cu2​O(s)+H2​(g)PH2{P_{H2}}PH2​ is the minimum partial pressure of H2{H_2}H2​(in bar) needed to prevent the oxidation at 1250K.1250K.1250K. The value of ln⁡(PH2)\ln \left( {{P_{H2}}} \right)ln(PH2​) is ‾\underline{\hspace{2cm}}​. Given: total pressure =1=1=1 bar, RRR(universal gas constant ) =8JK−1  mol−1,=8J{K^{ - 1}}\,\,mo{l^{ - 1}},=8JK−1mol−1,ln⁡(10)=2.3. \ln \left( {10} \right) = 2.3.\,ln(10)=2.3.Cu(s)Cu(s)Cu(s) and Cu2O(s)C{u_2}O\left( s \right)Cu2​O(s) are naturally immiscible. At 1250K:2Cu(s)1250K:2Cu(s)1250K:2Cu(s)+12O2(g)→Cu2O(s);+ \frac{1}{2}{O_2}\left( g \right) \to C{u_2}O\left( s \right);+21​O2​(g)→Cu2​O(s);ΔG∘=−78,000J mol−1\Delta {G^ \circ } = - 78,000J\,mo{l^{ - 1}}ΔG∘=−78,000Jmol−1H2(g)+12O2(g)→H2O(g);{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( g \right);H2​(g)+21​O2​(g)→H2​O(g);                  ΔG∘=−1,78,000J mol−1;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G^ \circ } = - 1,78,000J\,mo{l^{ - 1}};ΔG∘=−1,78,000Jmol−1; (GGG is the Gibbs energy)
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