Question

0.25 mol of CO and 0.45 mol of NO2 are mixed in a 25 L reaction vessel at 350°C. Calculate the molarities of all compounds present in the vessel once the equilibrium is reached. (6 points) CO(g) + NO2(g) ⇌ CO2(g) + NO(g) Kc = 67.5

Answer 1

Answer:

[CO₂] = [NO]  ≈ 0.11 mol/dm³

[CO]  = 0.25 mol/(25 L) = 0.01 mol/dm³

[NO₂] = 0.45 mol/(25 L) = 0.018 mol/dm³

Explanation:

The number of moles of CO = 0.25 moles

The number of moles of NO₂ = 0045 moles

The temperature of the vessel = 350°C

The equilibrium constant, Kc = 67.5

We have;

The product of the the concentration of the products = Kc × The product of the concentration of the reactants, therefore;

[CO₂] × [NO] = 67.5 ×(0.25/25 × 0.45/25) = 0.01215

Given that [CO₂] = [NO], we get;

[CO₂] = [NO] = √0.01215 ≈ 0.11 mol/dm³

[CO]  = 0.25 mol/(25 L) = 0.01 mol/dm³

[NO₂] = 0.45 mol/(25 L) = 0.018 mol/dm³