The free energy of formation of NO is 78 kJ mol –1 at the temperature of an automobile…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
The free energy of formation of NO is 78 kJ mol–1 at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at this reaction at 1000 K ?
\(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}(\mathrm{~g})\)
\(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}(\mathrm{~g})\)
- 8.4 × 10–5
- 7.1 × 10–9
- 4.2 × 10–10
- 1.7 × 10–19
Answer
(A) 8.4 × 10 –5
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