Thermodynamics C Question 4
Question 4 - 2024 (27 Jan Shift 2)
For a certain thermochemical reaction $\mathrm{M} \rightarrow \mathrm{N}$ at $\mathrm{T}=400 \mathrm{~K}, \Delta \mathrm{H}^{\ominus}=77.2 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}=122 \mathrm{JK}^{-1}, \log$ equilibrium constant $(\log \mathrm{K})$ is _______ $\times 10^{-1}$.
Show Answer
Answer (37)
Solution
$\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$
$=77.2 \times 10^{3}-400 \times 122=28400 \mathrm{~J}$
$\Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}$
$\Rightarrow 28400=-2.303 \times 8.314 \times 400 \log \mathrm{K}$
$\Rightarrow \log \mathrm{K}=-3.708=-37.08 \times 10^{-1}$
