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Question

A mixture of 3.8 moles of NO and 0.924 moles of CO2 was allowed to react in a 1.0L container. The equilibrium mixture contains 0.124 mole of CO2. Calculate Kc for the reaction:
NO(g)+CO(2(g)NO2(g)+CO(g).

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Solution

NO(g)+CO2(g)NO2(g)+CO(g)initialmoles:3.80.924Ateqm:3.80.8=30.9240.8=0.1240.80.8KC=Kn(1Vtotal)Δng=(0.8×0.83×0.124)=1.72

the correct answer is 1.72

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