(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 Kp = 0.100.
P4(g) Ìý ⇄ Ìý Ìý2 P2(²µ)Ìý
In an experiment, some P4 gas was added to an empty reaction vessel and then the vessel was quickly sealed. ÌýThe total pressure at equilibrium was 1.00 atm. ÌýWhat was the initial pressure of P4 used in this experiment?
Only 8% of the students writing the exam answered this question correctly. ÌýEvidence that suggests students regarded this question as tough, or perhaps just too laborious, is that 63% of students did not answer this question. Ìý
In a way, the question is a straightforward equilibrium problem that can be set up and solved using a tabular approach. ÌýLet me set up the problem in a way that, I think, most students would.ÌýÌý Ìý
| Reactants/Products | P4(g) | 2 P2(g) |
|---|---|---|
| Initial pressures | Ìý±ÊinitialÌý | 0 |
| Pressure changes | − x | + 2x |
| Equilibrium pressures | Pinitial − x | 2x |
In the summary above, Pinitial is the unknown initial pressure, and x is the amount by which the pressure of P4 has decreased when equilibrium has been reached. ÌýBecause the reaction produces 2 moles of P2 for every mole of P4 that reacts, the pressure of P2 has increased by 2x when equilibrium has been reached. ÌýAt equilibrium, the following conditions must be satisfied.
Ìý
Equation (1) is the equilibrium constant expression written in terms of the equilibrium partial pressures of P4 and P2. 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. ÌýWe can use the equation on the right to eliminate either x or Pinitial from the equation on the left, leaving us with one equation in one unknown. ÌýFor example, if we write Pinitial = 1.00 − x, and then substitute this expression for Pinitial 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. ÌýThe quadratic formula yields two roots, x = −0.185 and x = 0.135, but of course the positive root is the only physically meaningful one. Ìý(In setting up the table above, we assumed implicitly that x is a positive quantity.) ÌýIf we substitute x = 0.135 into equation (2), then we obtain
Pinitial = 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. ÌýIn one approach, we can set up the problem slightly differently and focus on the fraction of P4 that reacts. Ìý(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. ÌýFor 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.) ÌýLet αÌýrepresent the fraction of P4 that reacts. The equilibrium summary is as follows.ÌýÌý
| Reactants/Products | P4(g) | 2 P2(g) |
|---|---|---|
| Initial pressures | Ìý±ÊinitialÌý | 0 |
| Pressure changes | − αÌý±ÊinitialÌý | + 2αÌý±Êinitial |
| Equilibrium pressures | Pinitial(1−α) | 2α PinitialÌýÌý |
ÌýÌý ÌýÌýÌý ÌýÌýÌý
At equilibrium, the following conditions must be satisfied.
From equation (5), we get Pinitial = 1.00/(1 + α) and if we substitute this expression for Pinitial 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 αÌýinto equation (5) and solve for Pinitial, we obtain Pinitial = 0.865 atm.
We can avoid using the quadratic formula to solve equation (3) if we employ the method of successive approximations. ÌýWe can write equation (5) in the following form.
Because Kp is small, we expect that only a small amount of P2 will be formed. ÌýIf we assume that 2x is small compared to 1.00, then equation (8) yields
Ìý
Ìý
This is our first approximation for x. ÌýWe obtain a refined estimate of x by substituting x = 0.158 into the right-hand side of equation (8): Ìý
Ìý
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:ÌýÌý ÌýÌýÌý Ìýx = 0.136
4th approximation:ÌýÌý ÌýÌýÌý Ìýx = 0.135
5th approximation:ÌýÌý ÌýÌýÌý Ìýx = 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. ÌýOne might argue that it would have been easier to solve the original equation, equation (3), using the quadratic formula. ÌýI 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.
