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Gibbs Free Energy & Spontaneity

The standard ...

Question

The standard Gibbs free energy change $(Δ G^{o}$ ) at $25^{o} C$ for the dissociation of $N_{2} O_{4} (g)$ to ${N O}_{2} (g)$ is (given, equilibrium constant $= 0.15, R = 8.314 J K / m o l$ ):

1.1 k J

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4.7 k J

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8.1 k J

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38.2 k J

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Solution

The correct option is B $4.7 k J$
$Δ G^{o} = - 2.303 R T log K$

$= - 2.303 \times 8.314 \times 298 \times log 0.15$

$= - 2.303 \times 8.314 \times 298 \times (- 0.8239)$

$= 4700.25 J = 4.7 k J$

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