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)
Pн₂ is the minimum partial pressure of H₂ (in bar) needed to prevent the oxidation at 1250 K. The value of
ln (Pн₂) 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)

Answer 1

Answer:

no longer questions plz

sry dear