1. 2NO(g) + O2(g) -> 2NO2(g)
2. Let x be the change in concentration for NO2. Initially, the concentrations are:
[NO] = 4 mol / 1 L = 4 M
[O2] = 2.5 mol / 1 L = 2.5 M
[NO2] = 0 M
At equilibrium, [NO] = 4 - 2x M
[O2] = 2.5 - x M
[NO2] = 2x M
Plugging in the values to the equilibrium constant expression:
Kc = [NO2]^2 / ([NO] * [O2])
0.001 = (2x)^2 / ((4 - 2x) * (2.5 - x))
Solving for x, we get x = 0.401 M. Therefore, at equilibrium:
[NO] = 2.198 M
[O2] = 2.099 M
[NO2] = 0.802 M
3. Since the overall enthalpy is positive, the reaction is endothermic. Increasing the temperature will favor the formation of NO2 (product), therefore shifting the equilibrium to the right.
4. If the volume of NO2 is increased, the equilibrium will shift to the left in order to produce more NO and O2 which will reduce the increase in volume of NO2.
5. If the temperature is increased, the equilibrium constant (Kc) will change. Generally, for an endothermic reaction, increasing the temperature will increase the value of Kc.
6. If the concentration of NO is increased, the reaction will shift to the right to relieve the stress caused by the increased concentration of reactant. This will result in an increase in the concentration of NO2 and a decrease in the concentrations of NO and O2.
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 a balanced chemical equation for the above reaction?
2. Determine the equilibrium concentrations for each substance?
3. If the overall enthalpy of the above reaction is 49kJ/mol. Predict the outcome of the reaction?
4. If the volume of the product NO2 is being increased, where will equilibrium shift to and why will it have to shift that way?
5. What is happen to the equilibrium constant (Kc) value if temperature is increased?
6. If some concentration of Nitrogen oxide (NO) were increased, what would happen to the equilibrium?
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