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 1250K- 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 Cu s - H 2 O g - Cu 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 isGiven- total pressure -1 bar- R universal gas constant -8 JK -1 mol -1- ln 10-2-3 - Cu s and Cu 2 O s are mutually immiscible- At 1250 K - 2 Cu s -1-2 O 2 g - Cu 2 O s - - G -78-000 J mol -1H 2 g -1-2 O 2 g - H 2 O g - - G -1-78-000 J mol -1 - G is the Gibbs energy

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Equilibrium Constant and Standard Free Energy Change

The surface o...

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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)

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N_{2}

1250 K

N_{2}

1

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

ρ_{H_{2}}

H_{2}

1250 K

l n (ρ_{H_{2}})

= 1 b a r

R

= 8 J K^{- 1} m o l^{- 1}

l n (10) = 2.3

C u (s)

C u_{2} O (s)

1250 K

2 C u (s) + 1 / 2 O_{2} (g) \to C u_{2} O (s); Δ G^{\circ} = - 78, 000 J m o l^{- 1}

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

600 K

2 C u (s) + 1 / 2 O_{2} (g) ⇌ C u_{2} O (s); K_{p} = 7.5 \times 10^{10} {a t m}^{- 1 / 2}

O_{2}

X

10 X

2 C u O (s) ⟶ {C u}_{2} O (s) + \frac{1}{2} O_{2} (g)

Δ H = 144.6 k J {m o l}^{- 1}

Δ S = 0.116 k J {m o l}^{- 1} K^{- 1}

Δ G

N_{2} O_{4} (g)

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ N O (g)

{Δ G} = 86.6 k J

N O (g) + \frac{1}{2} O_{2} (g) ⟶ N O_{2} (g)

{Δ G} = - 34.82 k J

{2 N O}_{2} (g) ⟶ N_{2} O_{4} (g)

{Δ G} = - 5.77 k J

O_{2}

H_{2}

125.3 L

125^{\circ} C

H_{2} O (g)

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