### Thermodynamics Question 4

#### Question 4 - 2024 (27 Jan Shift 2)

For a certain thermochemical reaction $M \rightarrow N$ at $T=400 K, \Delta H^{\ominus}=77.2 kJ mol^{-1}, \Delta S=122 JK^{-1}, \log$ equilibrium constant $(\log K)$ is _______ $\times 10^{-1}$.

## Show Answer

#### Answer (37)

#### Solution

$\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$

$=77.2 \times 10^{3}-400 \times 122=28400 J$

$\Delta G^{\circ}=-2.303 RT \log K$

$\Rightarrow 28400=-2.303 \times 8.314 \times 400 \log K$

$\Rightarrow \log K=-3.708=-37.08 \times 10^{-1}$