1. The balanced chemical equation for the reaction is:
2NO(g) + O2(g) ⇌ 2NO2(g)
2. Let x be the change in concentration at equilibrium for both NO and O2.
Initial concentrations:
[NO] = 4 mol / (1000 ml) = 0.004 M
[O2] = 2.5 mol / (1000 ml) = 0.0025 M
[NO2] = 0 M (as it is produced)
At equilibrium:
[NO] = 0.004 - 2x
[O2] = 0.0025 - x
[NO2] = 2x
From the equilibrium constant expression:
Kc = [NO2]^2 / ([NO] * [O2])
0.001 = (2x)^2 / ((0.004 - 2x) * (0.0025 - x))
Solving the equation gives:
x = 0.0005 M
At equilibrium:
[NO] = 0.004 - 2*0.0005 = 0.003 M
[O2] = 0.0025 - 0.0005 = 0.002 M
[NO2] = 2*0.0005 = 0.001 M
3. Since the overall enthalpy of the reaction is positive (49 kJ/mol), the reaction is endothermic. This means that the forward reaction (2NO + O2 → 2NO2) will be favored at higher temperatures. The equilibrium position will shift towards the products, resulting in an increase in the concentration of NO2 at equilibrium.
In a chemical reaction 4mol of Nitrogen monoxide NO and 2.5mol of Oxygen O2 are mixed together in a container with a volume of 1000ml to produce Nitrogen Dioxide NO2. If equilibrium constant Kc for this reaction is 0.001.
1. Write the balance chemical equation for the above reaction?
2. Determine the equilibrium concentrations for each substances?
If the overall enthalpy of the above reaction is 49kJ/mol. Predict the outcome of the reaction?
1 answer
1 answer