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 oxidation at 1250 K. The value of ln ⁡ p H 2 is ____. (Given: total pressure = 1 bar, R (universal gas constant) = 8 J K - 1 m o l - 1 , ln ⁡ 10 = 2.3 . C u ( s ) a n d C u 2 O ( s ) are mutually immiscible. At 1250 K : 2 C u s + 1 2 O 2 g → C u 2 O ( s ) ; ∆ G ⊝ = - 78,000 J m o l - 1 H 2 g + 1 2 O 2 g → H 2 O ( g ) ; ∆ ⊝ = - 1,78,000 J m o l - 1 ; G is the Gibbs energy)

Question

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 oxidation at 1250 K. The value of ln ⁡ p H 2 is ____. (Given: total pressure = 1 bar, R (universal gas constant) = 8 J K - 1 m o l - 1 , ln ⁡ 10 = 2.3 . C u ( s ) a n d C u 2 O ( s ) are mutually immiscible. At 1250 K : 2 C u s + 1 2 O 2 g → C u 2 O ( s ) ; ∆ G ⊝ = - 78,000 J m o l - 1 H 2 g + 1 2 O 2 g → H 2 O ( g ) ; ∆ ⊝ = - 1,78,000 J m o l - 1 ; G is the Gibbs energy),

MARKS App · Accepted Answer

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

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