(This is a reprint from the October 2008 issue of Chem 13 News, pages 4-5.)

**At a certain temperature, the equilibrium constant for the reaction below is K _{p} = 0.100.**

**P _{4}(g) ⇄ 2 P_{2}(g) **

**In an experiment, some P _{4} gas was added to an empty reaction vessel and then the vessel was quickly sealed. The total pressure at equilibrium was 1.00 atm. What was the initial pressure of P_{4} used in this experiment?**

Only 8% of the students writing the exam answered this question correctly. Evidence that suggests students regarded this question as tough, or perhaps just too laborious, is that 63% of students did not answer this question.

In a way, the question is a straightforward equilibrium problem that can be set up and solved using a tabular approach. Let me set up the problem in a way that, I think, most students would.

Reactants/Products | P_{4}(g) |
2 P_{2}(g) |
---|---|---|

Initial pressures | P_{initial} |
0 |

Pressure changes | − x | + 2x |

Equilibrium pressures | P_{initial} − x |
2x |

In the summary above, P_{initial} is the unknown initial pressure, and x is the amount by which the pressure of P_{4} has decreased when equilibrium has been reached. Because the reaction produces 2 moles of P_{2} for every mole of P_{4} that reacts, the pressure of P_{2} has increased by 2x when equilibrium has been reached. At equilibrium, the following conditions must be satisfied.

Equation (1) is the equilibrium constant expression written in terms of the equilibrium partial pressures of P_{4} and P_{2}. Equation (2) results from the application of Dalton’s Law of partial pressures: the total pressure is equal to the sum of the partial pressures. We can use the equation on the right to eliminate either x or P_{initial} from the equation on the left, leaving us with one equation in one unknown. For example, if we write P_{initial} = 1.00 − x, and then substitute this expression for P_{initial} into equation (1), we obtain the following result.

This is a quadratic equation in x and, with appropriate rearrangement, it can be solved, using the quadratic formula, which was provided on the Data Sheet. The quadratic formula yields two roots, x = −0.185 and x = 0.135, but of course the positive root is the only physically meaningful one. (In setting up the table above, we assumed implicitly that x is a positive quantity.) If we substitute x = 0.135 into equation (2), then we obtain

P_{initial} = 0.865 atm.

Solving the quadratic equation using the quadratic formula is indeed laborious. In this particular case, there are two ways to avoid using the quadratic formula. In one approach, we can set up the problem slightly differently and focus on the fraction of P_{4} that reacts. (I advocate this approach because it allows me to emphasize, before setting up the equilibrium summary, that only a fraction of the reactants will be consumed in the reaction. For equilibria involving a weak acid or a weak base in aqueous solution, the fraction that reacts is related, in an obvious way, to the percentage ionization.) Let α represent the fraction of P_{4} that reacts. The equilibrium summary is as follows.

Reactants/Products | P_{4}(g) |
2 P_{2}(g) |
---|---|---|

Initial pressures | P_{initial} |
0 |

Pressure changes | − α P_{initial } |
+ 2α P_{initial} |

Equilibrium pressures | P_{initial}(1−α) |
2α P_{initial} |

At equilibrium, the following conditions must be satisfied.

From equation (5), we get P_{initial} = 1.00/(1 + α) and if we substitute this expression for P_{initial} into equation (4), we obtain the following equation.

Equation (6) above can be solved for α, as shown below.

If we substitute this value for α into equation (5) and solve for P_{initial}, we obtain P_{initial} = 0.865 atm.

We can avoid using the quadratic formula to solve equation (3) if we employ the method of successive approximations. We can write equation (5) in the following form.

Because K_{p} is small, we expect that only a small amount of P_{2} will be formed. If we assume that 2x is small compared to 1.00, then equation (8) yields

^{ }

This is our first approximation for x. We obtain a refined estimate of x by substituting x = 0.158 into the right-hand side of equation (8):

This is the second approximation. The process can be repeated until the desired accuracy is reached. The next three estimates of x are as follows.

3rd approximation: x = 0.136

4th approximation: x = 0.135

5th approximation: x = 0.135

After only a few quick calculations, we see that the value of x converges to a value of 0.135, which is the same value we obtained when we solved equation (3) using the quadratic formula. One might argue that it would have been easier to solve the original equation, equation (3), using the quadratic formula. I will not disagree, but I will point out that the method of approximations is more general, and thus more useful, because it can be used to solve more complicated equations (e.g., cubic equations) that are encountered occasionally when solving equilibrium problems.

In summary, I have presented two ideas that I hope you will consider using when solving equilibrium problems in the classroom.