MCQ
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_______ $\times 10^{-1}$.
  • $37$
  • B
    $38$
  • C
    $39$
  • D
    $40$

Answer

Correct option: A.
$37$
a
$ \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}$

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