# What is K if DeltaG^@ = -18.0 kJ for a reaction at 25 degrees C?

Apr 16, 2017

$\textsf{K = 6.97 \times {10}^{- 4}}$

#### Explanation:

$\textsf{\Delta {G}^{\circ} = - R T \ln K}$

$\therefore$$\textsf{\ln K = - \frac{\Delta {G}^{\circ}}{R T}}$

$\textsf{\ln K = - \frac{18 \times {10}^{3}}{8.31 \times 298} = - 7.268}$

$\textsf{K = 6.97 \times {10}^{- 4}}$