NEET Chemistry: Class 11, Chapter 7: Equilibrium questions with solutions

Q1. The plot shows the variation of \( -\ln \boldsymbol{K}_{\boldsymbol{p}} \) versus temperature for the two reactions. \( M(s)+\frac{1}{2} O_{2}(g) \rightarrow M O(s) \) and \( C(s)+\frac{1}{2} O_{2}(g) \rightarrow C O(s) \) Identify the correct statement?

1. \( A t T>1200 K, \) carbon will reduce \( M O(s) \) to \( M(s) \)
2. At \( T<1200 \mathrm{K} \), oxidation of carbon is unfavourable.
3. At \( T<1200 \) K, the reaction \( M O(s)+C(s) \rightarrow M(s)+ \) \( C O(g) \) is spontaneous
4. Oxidation of carbon is favourable at all temperatures

Answer: \( A t T>1200 K, \) carbon will reduce \( M O(s) \) to \( M(s) \)

At temperatures above 1200 K, the reaction becomes spontaneous, and carbon can reduce MO(s) to M(s). [AI-generated key — verify before high-stakes use]