WebLecture

- Determining the Equilibrium Constant
- Using the Equilibrium Constant
- Practice with Concepts
- Discussion Questions
- Optional Website Reading

Purpose | Formula | Variables | ||

Equilibrium Constant (Q at equilibrium) |
| [C] and [D]: concentrations of products [A] and [B]: concentrations of reactants | ||

In order to determine a reaction *rate* we need to experimentally determine the change in concentration as a function of time. We looked at how to do this in preceding chapters. Now we want to determine the equilibrium constant by looking at the state of a reaction system at the start of the reaction and comparing it to the state of the system when the reaction has reached equilibrium, and no further changes are taking place in the concentration of the reactants and products.

To do this, we need the concentrations at the start and equilibrium point of the reaction. The easiest and most clear way to look at this data is to set it out the table that includes the initial conditions (I), the change in reactant and product concentrations (C), in the equilibrium conditions (E), a method sometimes referred to as making an "ICE" table.

Consider the sample reaction 2A + B → 2C, where A and B are reactants that combine to form the product C. Our table will need to show the changes in [A] ( where the brackets indicate concentration in moles per liter), [B] and [C]. We need to remember that for every two moles of A, one mole of B will be consumed, and to moles of C will be created. so a change of X amount in B will require a change of 2X amount in A and C. If we use X for the change amounts, we can represent the final state of the system using initial conditions.

Equation | 2A | + B | ↔ C |

Initial concentrations (mol/L) | 0.01 | 0.02 | 0.00 |

Change Amounts (mol/L) | -2X | -X | +2X |

Equilibrium concentrations (mol/L) | 0.01-2X | 0.02-X | 0.00+2X |

The equilibrium constant in this case will be

$$K=\frac{{\mathrm{[C]}}^{2}}{{\mathrm{[A]}}^{2}\mathrm{[B]}}=\frac{{\mathrm{(0.00\; +\; 2X)}}^{2}}{{\mathrm{(0.01\; -\; 2X)}}^{2}\mathrm{(0.02\; -X)}}$$

We can determine K then experimentally by measuring X in any one of the species involved (A, B, or C), although usually we would measure concentration changes for all three in order to reduce error. Suppose that experimentally we discovered that the concentrations change by 0.004*M* for B, and therefore twice that much, or 0.008*M* for A and C. Our calculation of K, based on experimental data, becomes:
$$K=\frac{{\mathrm{(0.00\; +\; 0.008)}}^{2}}{{\mathrm{(0.01\; -\; 0.008)}}^{2}\mathrm{(0.2\; -\; 0.004)}}=\frac{{\mathrm{(0.008)}}^{2}}{{\mathrm{(0.002)}}^{2}\mathrm{(0.196)}}=\frac{\mathrm{(0.000064)}}{\mathrm{(0.000004)}\mathrm{(0.196)}}=\mathrm{81.632}$$

Remember that an equilibrium constant greater than one indicates a product favored reaction, so in this case we would expect the reaction to proceed nearly to completion, with most of the reactants used up.

The equilibrium constant equation as we presented it above has three sets of variables, then: the initial concentrations, the amount of change, and the value of K. If we know any two of the sets, we can determine the third set of values. So once we know K from experimental data for one set of initial concentrations, we can predict the change X for a different set of initial concentrations.

If we have the reaction

aA + bB → cC + Dd

then the amount of change will be proportional to the stoiochiometric amounts identified by a, b, c, and d, and the equilibrium constant is given by the general equation. We can rewrite the amounts of change for each of the other species in terms of the change for one of the species, e.g., ΔB = b/a ΔA. This allows us in most cases to predict x if K has been experimentally determined.

$$K=\frac{{\mathrm{[C]}}^{\mathrm{c}}{\mathrm{[D]}}^{\mathrm{d}}}{{\mathrm{[A]}}^{\mathrm{a}}{\mathrm{[B]}}^{\mathrm{b}}}=\frac{{\mathrm{({C}_{0}+\; \Delta C)}}^{\mathrm{c}}{\mathrm{({D}_{0}+\; \Delta D)}}^{\mathrm{d}}}{{\mathrm{({A}_{0}-\; \Delta A)}}^{\mathrm{a}}{\mathrm{({B}_{0}-\; \Delta B)}}^{\mathrm{b}}}=\frac{{\mathrm{({C}_{0}+\; c/a\Delta A)}}^{\mathrm{c}}{\mathrm{({D}_{0}+\; d/a\Delta A)}}^{\mathrm{d}}}{{\mathrm{({A}_{0}-\; \Delta A)}}^{\mathrm{a}}{\mathrm{({B}_{0}-\; b/a\Delta A)}}^{\mathrm{b}}}=\frac{{\mathrm{({C}_{0}+\; (c/a)x)}}^{\mathrm{c}}{\mathrm{({D}_{0}+\; (d/a)x)}}^{\mathrm{d}}}{{\mathrm{({A}_{0}-\; x)}}^{\mathrm{a}}{\mathrm{({B}_{0}-\; (b/a)x)}}^{\mathrm{b}}}$$Suppose that we know that our equilibrium constant is 81.632 from an earlier experiment. We want to use a different set of initial concentration values.

Equation | 2A | + B | ↔ C |

Initial concentrations (mol/L) | 0.15 | 0.075 | 0.00 |

Change Amounts (mol/L) | -2X | -X | +2X |

Equilibrium concentrations (mol/L) | 0.15-2X | 0.075-X | 0.00+2X |

Now we have

$$K=\mathrm{81.632}=\frac{{\mathrm{(0.00\; +\; 2X)}}^{2}}{{\mathrm{(0.15\; -\; 2X)}}^{2}\mathrm{(0.075\; -\; X)}}$$

We can solve this particular equation approximately, if X is very little (unlikely in this case), or with difficulty, since it yields a cubic equation: $$\frac{{\mathrm{(2X)}}^{2}}{K}={\mathrm{(0.15\; -\; 2X)}}^{2}\mathrm{(0.075\; -\; X)}$$

$$\frac{{\mathrm{4X}}^{2}}{K}=({\mathrm{0.15}}^{2}+4{X}^{2}-0.6\mathrm{X)}\mathrm{(0.075\; -\; X)}$$

$$\frac{{\mathrm{4X}}^{2}}{K}=0.016875+0.3{X}^{2}-0.045X-0.0225X-4{X}^{3}+0.6{X}^{2}$$

$$-\mathrm{4K}{X}^{3}-\mathrm{3.1K}{X}^{2}-\mathrm{0.0675K}X+\mathrm{0.016875K}=0$$

$$-326.528{X}^{3}-253.059{X}^{2}-5.51016X+0.137754=0$$

This looks unappetizing, but it is solvable. Because we are chemists and not mathmaticians, we'll resort to WolramAlpha's cubic equation solver, which gives an X value of 0.751807*M*.

In many cases, X is very small, and we can ignore factors in X^{3}.

- Why is predicting the outcome of a chemical reaction important?

Calculating Equilibrium Constants provides another tutorial page with examples for different situations.

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