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 $l n (P_{H_{2}})$ is

(Given: total pressure =1 bar, R (universal gas constant) $= 8 J K^{- 1} m o l^{- 1}$ , ln(10)=2.3. Cu(s) and $C u_{2} O (s)$ are mutually immiscible.
At $1250 K : 2 C u (s) + \frac{1}{2} O_{2} (g) \to C u_{2} O (s); Δ G^{\circ} = - 78, 000 J m o l^{- 1}$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ G^{\circ} = - 1, 78, 000 J m o l^{- 1}; G$ 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) \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

l n (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. Cu(s) and

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); Δ G^{\circ} = - 78, 000 J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ G^{\circ} = - 1, 78, 000 J m o l^{- 1}; G

is the Gibbs energy)

Toppr Admin · Accepted Answer