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Compute the equilibrium constant at 25 °C for the following reaction: SnO2(s) + 2 CO(g) ...

Compute the equilibrium constant at 25 °C for the following reaction:

SnO2(s) + 2 CO(g)  2 CO2(g) + Sn(s, white)

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Answer #1

SnO2(s) + 2 CO(g) -------> 2 CO2(g) + Sn(s, white) : ΔG = >

ΔG = ΔG0fproducts – ΔG0freactants

ΔG = [(2x ΔG0fCO2(g) )+ ( ΔG0fSn(s) )]- [( 2 x ΔG0fCO(g )+ ( ΔG0fSnO2 (s) )]

    = [(2x(-386.01)+0] - [(2x(-137.3))+ (-519.65)] kJ/mol

    = +22.23 kJ

Also ΔG = -RT lnK

Where R = gas constant = 8.314x10-3 kJ/mol-K

T = Temperature = 25 oC = 25 + 273 = 298 K

K = Equilibrium constan t = ?

Plug the values we get ln K = - ΔG /(RT) = -8.97

K = e-8.97 = 1.27x10-4

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