Answer
$K_{eq}= 4.6×10^3$
Work Step by Step
Consider the formula for calculating the equilibrium constant :
$K_{eq}=\frac{([C]_eq^c [B]_eq^a)}{([A]_eq^a [B]_eq^b )}=e^{(-∆°G')/RT}$
We know the following values :
$∆G°^{'}=-20.9 kJ∙mol^{-1}$
$R= 8.314 J∙ mol^{-1}∙K^{-1}$
$T=25℃ $
Before we can proceed with the calculation, we need to first make sure that the units match up.
$T=25℃$ needs to be converted to Kelvin $→$ $T (K)= T(℃)+273$ $→$ $ T(K)=25+273=298$
$∆G°^{'}= -20.9 kJ∙mol^{-1}$ needs to be converted to $kJ∙mol^{-1}$
$\frac{-20.9 kJ}{mol} ∙\frac{1000 J}{1 kJ}=-20.9 J∙ mol^{-1} $
Now
$ T= -248 K$
$R= 0.008314 kJ∙ mol^{-1}∙K^{-1}$
$∆G°^{'}=-20.9 kJ∙mol^{-1}$
and
$K_{eq}=e^{(-∆°G)/RT}$
$=e^{\frac{-(-20.9 ×10^3 J∙mol^{-1})}{8.314 J∙ mol^{-1}∙K^{-1} ∙ (298 K)}}$
$=e^{8.4357} = 4.6 \times 10^{3}$
Thus,
$K_{eq}=4.6 \times 10^{3}$
