JEE Advanced2018ChemistryThermodynamics (C)Actual
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: 2 C u s + H 2 O g → C u 2 O ( s ) + H 2 ( g ) P H 2 Is the minimum partial pressure of H 2 (in bar) needed to prevent the oxidati
Correct answer
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Step-by-step solution
2 C u s + 1 4 O 2 g → 1 C u 2 O s ∆ G 0 = - 78 k J H 2 g + 1 2 O 2 → H 2 O g ∆ G o = - 178 k J × - 1 Hence, 2 C u s + H 2 O g → C u 2 O + H 2 g ∆ G o = + 100 k J ∆ G = ∆ G o + R T ln Q 0 = + 100 + 8 1000 × 1250 ln P H 2 P H 2 O - 100 × 1000 8 = 1250 ln P H 2 1 100 × 1 ln P H 2 = - 14.6