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Problems - Reaction Equilibrium

Author: John Hutchinson

Problem 1 In the data given for equilibrium of this reaction, there is no volume given. Show that changing the volume for the reaction does not change the number of moles of reactants and products present at equilibrium, i.e. changing the volume does not shift the equilibrium.
Problem 2 For this reaction the number of moles of NO2 at equilibrium increases if we increase the volume in which the reaction is contained. Explain why this must be true in terms of dynamic equilibrium, give a reason why the rates of the forward and reverse reactions might be affected differently by changes in the volume.
Problem 3 We could balance equation 2 by writing

2N2(g)+6H2(g) 4NH3(g) [12]

Write the form of the equilibrium constant for the reaction balanced as in equation 12. What is the value of the equilibrium constant? (Refer to table 3.) Of course, the pressures at equilibrium do not depend on whether the reaction is balanced as in equation 2 or as in equation 12. Explain why this is true, even though the equilibrium constant can be written differently and have a different value.

Problem 4 Show that the equilibrium constant Kp in equation 8 for this reaction can be written in terms of the concentrations or particle densities, e.g. [N2]=nN2 V, instead of the partial pressures. In this form, we call the equilibrium constant Kc. Find the relationship between Kp and Kc, and calculate the value of Kc.
Problem 5 For each of these reactions, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants:

5.a) 2 CO(g)+O2(g) 2CO2(g)
5.b) O3(g) + NO(g) NO2(g)+O2(g)
5.c) 2 O3(g) 3 O2(g)

Problem 6 Plot the data in table 4 on a graph showing Kp on the y-axis and T on the x-axis. The shape of this graph is reminiscent of the graph of another physical property as a function of increasing temperature. Identify that property, and suggest a reason why the shapes of the graphs might be similar.
Problem 7  Using Le Châtelier's principle, predict whether the specified "stress" will produce an increase or a decrease in the amount of product observed at equilibrium for the reaction: 2H2(g) + CO(g) CH3OH(g) (13) ΔH°=-91kJ mol

7.a) Volume of container is increased.
7.b) Helium is added to container.
7.c) Temperature of container is raised.
7.d) Hydrogen is added to container.
7.e) CH3OH is extracted from container as it is formed.

Last Update: 2011-02-16