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 w

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:

$2 C u (s) + H_{2} O (g) \to C u_{2} O (s) + H_{2} (g)$

$P_{H_{2}}$ 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 $1250 K : 2 C u (s) + \frac{1}{2} O_{2} (g) \to C u_{2} O (s); ∆ G^{⊝} = - 78,000 J m o l^{- 1}$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); ∆^{⊝} = - 1,78,000 J m o l^{- 1}; G$ is the Gibbs energy)

TestHub · 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

Chemistry - Thermodynamics - II Question with Solution | TestHub

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