Calculate the equilibrium constant at 25 degrees celsius given the Standard Free Energy value of - 107.2 kJ

- 43.2

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43.2

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6.18 x

10^{8}

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1.04

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6.18 x

10^{9}

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Solution

The correct option is C 6.18 x $10^{8}$
The temperature is 25 deg C or 298 K
$Δ G^{0} = - R T l n K$
$Δ G^{0} = - 107.2 k J = - 107200 J$
$- 107200 = - 8.314 \times 298 \times l n K$
$l n K = 43.268$
$K = 6.18 \times 10^{18}$

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