In the mixture of NO and CO2 (initially containing 4 mol of NO and 0.9 mole of CO2) reaction occurs according to the equation below: NO(g) + CO2(g) ⇌ NO2(g) + CO(g). At equilibrium 0.1 mol of CO2 was present. What is the equilibrium constant at the temperature of this experiment?
Answer: 0.2
Explanation
Given reaction:
NO(g) + CO₂(g) ⇌ NO₂(g) + CO(g)
Initial moles:
NO = 4 mol
CO₂ = 0.9 mol
NO₂ = 0
CO = 0
At equilibrium:
CO₂ = 0.1 mol
→ So, change in CO₂ = 0.9 - 0.1 = 0.8 mol consumed
→ Therefore, 0.8 mol of NO also reacted
→ 0.8 mol each of NO₂ and CO formed
Equilibrium moles:
NO = 4 - 0.8 = 3.2 mol
CO₂ = 0.1 mol
NO₂ = 0.8 mol
CO = 0.8 mol
Equilibrium expression:
K = [NO2][CO]/[NO][CO2] = (0.8)(0.8)/(3.2)(0.1) = 0.64/0.32 = 0.2
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