JEE Main 2016 · 9 Apr · Shift 1 · Chemistry Q42

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ChemistrySingle correct+4 / -1Question 42 of 90

The plot shows the variation of −ln $K_p$ versus temperature for the two reactions.

$\ce{M(s) + 1/2O2(g) -> MO(s)}$ and $\ce{C(s) + 1/2O2(g) -> CO(s)}$

Identify the correct statement :

−ln Kp against T (K): the M → MO curve falls steeply and crosses the flatter C → CO curve at 1200 K (−ln Kp = 20); below 1200 K the C → CO curve is lower, above it the M → MO curve is lower.

At T > 1200 K, carbon will reduce MO(s) to M(s).

At T < 1200 K, the reaction

\ce{MO(s) + C(s) -> M(s) + CO(g)}

is spontaneous.

At T < 1200 K, oxidation of carbon is unfavourable.

Oxidation of carbon is favourable at all temperatures.

Answer and worked solution

Official answer: B

A lower −ln Kp means a larger Kp. Below 1200 K the C → CO curve is lower, so carbon is more readily oxidised than M and can reduce MO: $\ce{MO + C -> M + CO}$ is spontaneous.

Topics:Extraction and thermodynamics (Ellingham)

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JEE Main 2016 · 9 Apr · Shift 1, Chemistry question 42