Question

The plot shows the variation of \(-\ln K_p\) versus temperature for the two reactions \(\mathrm{M(s)}+\frac{1}{2}\mathrm{O_2(g)}\rightarrow\mathrm{MO(s)}\) and \(\mathrm{C(s)}+\frac{1}{2}\mathrm{O_2(g)}\rightarrow\mathrm{CO(s)}\).

Identify the correct statement :

• A. At T>1200 K, carbon will reduce MO(s) to M(s).
• B. At T<1200 K, the reaction \(\mathrm{MO(s)+C(s)\rightarrow M(s)+CO(g)}\) is spontaneous.
• C. At T<1200 K, oxidation of carbon is unfavourable.
• D. Oxidation of carbon is favourable at all temperatures.

Hint:

\(-\ln K_p\) is proportional to \(\Delta G/RT\); the lower curve is the more favourable oxidation.

Answer 1

The Correct Option is B

\(\Delta G^\circ=-RT\ln K_p\), so a lower \(-\ln K_p\) means a more negative \(\Delta G\) (more favourable oxidation).
Graph: for T < 1200 K the C to CO curve lies below the M to MO curve, so \(\Delta G_{C\to CO}\lt\Delta G_{M\to MO}\). For T > 1200 K the M to MO curve is lower.
For \(\mathrm{MO+C\rightarrow M+CO}\): \(\Delta G=\Delta G_{C\to CO}-\Delta G_{M\to MO}\), which is negative for T < 1200 K, so the reaction is spontaneous below 1200 K.
Hence (1) is wrong (above 1200 K carbon cannot reduce MO), (3) is wrong (oxidation of C is favourable below 1200 K), (4) is wrong (above 1200 K M oxidises more readily).
Answer: (2) At T<1200 K, the reaction MO(s) + C(s) \(\rightarrow\) M(s) + CO(g) is spontaneous.