GATE 2022 MT – Question 48

Metallurgical Thermodynamics · Reaction equilibria and Ellingham diagrams · 2 marks · Numerical answer

Find the equilibrium ratio $p_{CO}/p_{CO_2}$ at 1873 K for the reaction $\mathrm{Mo}(s)+\mathrm O_2(g)\rightleftharpoons\mathrm{MoO}_2(s)$ (two decimal places).

Given: $\Delta_fG^\circ_{1873}=-262300$ J, $a_{\mathrm{MoO}_2}=0.5$, and $\Delta_rG^\circ_{1873}=-120860$ J for $\mathrm{CO}(g)+0.5\mathrm O_2(g)\rightleftharpoons\mathrm{CO}_2(g)$. Use $R=8.314$ J K$^{-1}$ mol$^{-1}$.

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Correct answer: 2.7 to 2.76

Explanation

For unit Mo activity, metal–oxide equilibrium gives $p_{O_2}=0.5\exp[-262300/(RT)]$. CO oxidation gives $p_{CO_2}/p_{CO}=\exp[120860/(RT)]\sqrt{p_{O_2}}$. Inverting and substituting yields $p_{CO}/p_{CO_2}=\sqrt2\exp[(131150-120860)/(8.314\times1873)]=\boxed{2.74}$.