Chemistry · Delta G° (Standard Gibbs energy change) and equilibrium constant
The plot shows the variation of versus temperature for the two reactions. +\frac
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?
- A. \( A t T>1200 K, \) carbon will reduce \( M O(s) \) to \( M(s) \)
- B. At \( T<1200 \mathrm{K} \), oxidation of carbon is unfavourable.
- C. At \( T<1200 \) K, the reaction \( M O(s)+C(s) \rightarrow M(s)+ \) \( C O(g) \) is spontaneous
- D. Oxidation of carbon is favourable at all temperatures
Step-by-step solution
The plot of -ln Kp vs T is related to Gibbs free energy: -ln Kp = ΔG°/(RT). For the two reactions, the lines intersect at 1200 K, indicating equal Kp values. Above 1200 K, the line for C(s) + 1/2 O2(g) → CO(g) lies below that for M(s) + 1/2 O2(g) → MO(s), meaning -ln Kp is smaller, so Kp is larger for carbon oxidation. This implies that CO formation is more favorable, and the net reaction MO(s) + C(s) → M(s) + CO(g) becomes spontaneous (ΔG° < 0). Thus, carbon reduces MO to M above 1200 K.
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