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For the equil...

Question

For the equilibrium $N i O_{(s)} + C O_{(g)} ⇌ N i_{(s)} + C O_{2 (g)}, Δ G^{0} ({J mol}^{- 1}) = - 20, 700 - 11.97 T$ . Calculate the temperature (in K) at which the product gases at equilibrium at $1 atm$ will contain $400 ppm$ (parts per million) of carbon monoxide:( upto two decimal where $R = 8.314 {J K}^{- 1} {mol}^{- 1}$ )

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Solution

For the given reaction
$p_{C O} << p_{C O_{2}}$ , hence
$K_{p} = \frac{1}{p_{C O}} = \frac{1}{400 \times 10^{6}} = 2500$
$Δ G^{0} = - R T ln K_{p}$ ;
$ln K_{p} = \frac{Δ G^{0}}{R T} = \frac{20, 700 + 11.97 T}{R T}$
The equation when solved for $T$ using $R = 8.314 {J K}^{- 1} {mol}^{- 1}$ , gives $T = 390.27 K$

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Δ G = Δ G^{\circ} + R T l n Q_{P}

Δ G = 0 and Q_{p} = K_{P}

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1 M or 1 atm

Δ G^{\circ} = - v e or K > 1 : forward reaction is feasible

Δ G^{\circ} = + v e or K < 1 : reverse reaction is feasible

Δ G^{\circ} = 0 or K = 1 : reaction is at equilibrium (very rare)

N i O (s) + C O (g) ⇌ N i (s) + C O_{2} (g), Δ G^{\circ} ({J Mol}^{- 1}) = - 20700 - 11.97 T .

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400 ppm

10

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C O (g) + (\frac{1}{2}) O_{2} (g) \to C O_{2} (g); △ H^{o} = - 282.84 kJ/mol

S_{C O_{2} (g)}^{o} = 213.8 {J K}^{- 1} {mol}^{- 1}, S_{C O (g)}^{o} = 197.9 {J K}^{- 1} {mol}^{- 1}, S_{O_{2}}^{o} = 205.0 {J K}^{- 1} {mol}^{- 1}

△ G^{⊖}

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X (s) ⇌ Y (s) + Z (g)

l n \frac{p_{z}}{p^{0}} versus \frac{10^{4}}{T}

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\frac{d (l n K)}{d (\frac{1}{T})} = - \frac{Δ H^{0}}{R}

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