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Fall 2010 UMass Boston
CHEMICAL EQUILIBRIUM (ICE METHOD)
Introduction
- Chemical equilibrium occurs when opposing reactions are proceeding at equal rates.
- The rate at which the products are formed from the reactants equals the rate at which the reactants are formed from the products.
- As a result, concentrations cease to change, making the reaction appear to be stopped. How fast a reaction reaches this equilibrium state is a matter of kinetics.
- An equilibrium state results when a reaction is reversible
At equilibrium the concentrations of reactants and products is still changing, however, the rate of the forward reaction (kf ) is equal to the rate of the reverse reaction (kr ) in what is described as a dynamic equilibrium such that no change in their concentrations is observed. Thus, for equilibrium to occur, neither reactants nor products can escape from the system.
- The law of mass action states the ratio of forward and reverse processes is described by the equilibrium constant K C which can be calculated using a knowledge of the equilibrium concentrations of reactants and products.
The equilibrium constant expression depends only on the stoichiometry of the reaction, not on the mechanism.
Objectives and Success Criteria
- Mastering the application of the ICE table methodology to equilibrium problems.
- Accurate solutions to problems involving reactant and product concentrations and equilibrium constants.
Prerequisites
Fall 2010 UMass Boston
MODEL 1: The ICE Table
A worked example: Initially 1.50 moles of N 2 (g) and 3.50 moles of H 2 (g) were added to a 1 L container at 700 °C. As a result of the reaction
the equilibrium concentration of NH 3 (g) became 0.540 M. What is the value of the equilibrium constant for this reaction at the given temperature of 700 °C.
1.50 mol L‐^1 3.50 mol L‐^1 0 mol L‐^1
−x −3x +2x
1.50 mol L‐^1 3.50 mol L‐^1 +2x mol L‐^1 –x −3x
We are now set to solve for the equilibrium constant K (^) C using the equilibrium equation :
eqn. 1
where the reactant and product concentrations should be expressed at equilibrium. The problem tells us that the equilibrium concentration of NH 3 is 0.540 M, thus we can solve for the unknown ‘x’
[NH 3 ] eq = +2x = 0.540 M x = 0.270 M we can now solve for K C
N 2 (g) +^ 3H 2 (g) 2NH (^3)
I. Write the I nitial concentrations of reactants and products:
C. Write the C hange in concentration due to reaction using the given reaction stoichiometric coefficients:
E. Write the reactant and product concentrations at E quilibrium.
Fall 2010 UMass Boston
EXERCISES
- Initially, 1.0 mol of NO(g) and 1 mol of Cl 2 (g) were added to a 1 L container. As a result of the reaction
the equilibrium concentration of NOCl(g) became 0.96 M. Using the RICE table methodology determine the value of the equilibrium constant K C for this reaction.
Solve for K C :
Answer: K C = 1.11 x 103
I. Write the I nitial concentrations of reactants and products:
C. Write the C hange in concentration due to reaction using the given reaction stoichiometric coefficients:
E. Write the reactant and product concentrations at E quilibrium.
Fall 2010 UMass Boston
PROBLEMS
- If a 10.00 L flask at 500 K is filled with a 0.30 mole of hydrogen and 0.30 mole of iodine, what are the equilibrium concentrations of the three gases?
The equilibrium constant K C = 45.0. The relevant reaction is
Answer: [H 2 ] = [I 2 ] = 0.0069 M; [HI] = 0.0462 M
Fall 2010 UMass Boston
- Gaseous carbon dioxide is partially decomposed according to the following equation.
An initial pressure of 1.00 atm of CO 2 is placed in a closed container at 2500 K, and 2.1 % of the molecules decompose. Determine the equilibrium constant K p at this temperature.
Answer : K p = 4.83 x 10 ‐^6
