The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: 2Cu(s) + H2O(g) → Cu2O(s) + H2(g) pH2 is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln (pH2) is . (Given: total pressure = 1 bar, R (universal gas constant) = 8 J K−1 mol−1, ln(10) = 2.3. Cu(s) and Cu2O(s) are mutually immiscible. At 1250 K: 2Cu(s) + ½ O2(g) → Cu2O(s); ΔGƟ = − 78,000 J mol−1 H2(g) + ½ O2(g) → H2O(g); ΔGƟ = − 1,78,000 J mol−1; G is the Gibbs energy)

Question

The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: 2Cu(s) + H2O(g) → Cu2O(s) + H2(g) pH2 is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln (pH2) is . (Given: total pressure = 1 bar, R (universal gas constant) = 8 J K−1 mol−1, ln(10) = 2.3. Cu(s) and Cu2O(s) are mutually immiscible. At 1250 K: 2Cu(s) + ½ O2(g) → Cu2O(s); ΔGƟ = − 78,000 J mol−1 H2(g) + ½ O2(g) → H2O(g); ΔGƟ = − 1,78,000 J mol−1; G is the Gibbs energy) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 Cu ( s )+(1/4) O 2( g ) arrow 1 Cu 2 O ( s ) Δ G 0=-78 kJ [ H 2( g )+(1/2) O 2 arrow H 2 O ( g ). .Δ G0=-178 kJ ] ×(-1) Hence, 2 Cu ( s )+ H 2 O ( g ) arrow Cu 2 O + H 2( g ) Δ G 0=+100 kJ 0=+100+(8/1000) × 1250 ln (PH2/PH2 O ) -(100 × 1000/8)=1250 ln (PH2/((1)100 × 1)) ln PH=-14.6

2Cu(s)+41​O2​(g)→1Cu2​O(s) ΔG0=−78kJ [H2​(g)+21​O2​→H2​O(g) ΔG0=−178kJ]×(−1) Hence, 2Cu(s)+H2​O(g)→Cu2​O+H2​(g) ΔG0=+100kJ 0=+100+10008​×1250lnPH2​O​PH2​​​ −8100×1000​=1250ln(1001​×1)PH2​​​ lnPH​=−14.6