Question

What is the value of log K for the reaction at $298K$ if $\Delta {G^0} = 63.3kJ$ ?
A) $ - 11.09$
B) $ - 9.6$
C) $ - 7.4$
D) $ - 10.5$

Answer 1

Hint: $\Delta {G^0}$ stands for the standard Gibbs free energy for a reaction, with everything in the standard states (gases at $1$ bar, and solutions at $1$ M concentration), and at a specific temperature (usually$25^\circ C$ ). To calculate the value of log K, we need to find out the formula depicting the relationship between $\Delta {G^0}$ and log K and then figure out the value of log K.

Complete answer:
Now, the relationship between $\Delta {G^0}$ and log K is given in the following formula - $\Delta {G^0} = - 2.303RTlog{K_{sp}}$
Now we are provided with the value of $\Delta {G^0} = 63.3kJ$ and temperature of $298K$, so the calculation for log K will be like –
$\eqalign{
  & 63.3kJ = 63300J = - (2.303)(8.314)(298)logK \cr
  & logK = - 11.09 \cr} $
So, the calculated value of log K is found to be – ($ - 11.09$).
Hence, the correct option is A) $ - 11.09$.

Note:
Gibbs free energy, G, is used to describe the spontaneity of a process and the expression for it is - $G = H - T\Delta S$ . The free energy change $\Delta G$ is equal to $ - T\Delta {S_{univ}}$ and applies just to a system itself, without regard for the surroundings. It is defined by the Gibbs equation: $\Delta G = \;\Delta H - T\Delta S$
For a spontaneous process at constant temperature and pressure, $\Delta G$ must be negative. In many cases, we can predict the sign of $\Delta G$ from the signs of $\Delta H$ and $\Delta S$ .