# 13.7: Predicting the Direction of a Reaction

- Page ID
- 49524

Often you will know the concentrations of reactants and products for a particular reaction and want to know whether the system is at equilibrium. If it is not, it is useful to predict how those concentrations will change as the reaction approaches equilibrium. A useful tool for making such predictions is the **reaction quotient**, *Q*. *Q* has the same mathematical form as the equilibrium-constant expression, but *Q* is a ratio of the actual concentrations (not a ratio of equilibrium concentrations).

For example, suppose you are interested in the reaction

\[2 \text{SO}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{SO}_{3} (g)\]

\[ K_{\text{c}} = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = 245 \text{ mol/L} \text{(at 1000 K)} \]

and you have added 0.10 mol of each gas to a container with volume 10.0 L. Is the system at equilibrium? If not, will the concentration of SO_{3} be greater than or less than 0.010 mol/L when equilibrium is reached? You can answer these questions by calculating *Q *and comparing it with *K*_{c}. There are three possibilities:

- If
*Q*=*K*_{c}then the actual concentrations of products (and of reactants) are equal to the equilibrium concentrations and the system is at equilibrium. - If
*Q*<*K*_{c}then the actual concentrations of products are less than the equilibrium concentrations; the forward reaction will occur and more products will be formed. - If
*Q*>*K*_{c}then the actual concentrations of products are greater than the equilibrium concentrations; the reverse reaction will occur and more reactants will be formed.

For the reaction given above,

\[ Q = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = \frac{(\frac{0.1 mol}{10 L})^{2}}{(\frac{0.1 mol}{10 L})^{2} (\frac{0.1 mol}{10 L})} = 100 \frac{\text{mol}}{\text{L}} \]

(In the expression for *Q* each actual concentration is enclosed in braces {curly brackets} in order to distinguish it from the equilibrium concentrations, which, in the *K*_{c} expression, are enclosed in [square brackets].) In this case *Q *= 100 mol/L. This is less than *K*_{c}, which has the value 245 mol/L. This implies that the concentrations of products are less than the equilibrium concentrations (and the concentrations of reactants are greater than the equilibrium concentrations). Therefore the reaction will proceed in the forward direction, producing more products, until the concentrations reach their equilibrium values.

Example \(\PageIndex{1}\): Equilibrium

At 2300 K, the equilibrium constant, *K*_{c}, is 1.7 x 10^{-3} for the reaction

\[ \text{N}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{NO} (g) \]

A mixture of the three gases at 2300 K has these concentrations, [N_{2}] = 0.17 mol dm^{-3}, [O_{2}] = 0.17 mol dm^{-3}, and [NO] = 0.034 mol dm^{-3}.

(a) Is the system at equilibrium?

(b) In which direction must the reaction occur to reach equilibrium?

(c) What are the equilibrium concentrations of N_{2}, O_{2}, and NO?

**Solution**

Use the known concentrations to calculate *Q*. Compare *Q* with *K*_{c} to answer questions (a) and (b). Use an ICE table to answer part (c).

\[Q = \frac{\{\text{NO\}}^{2}}{\{\text{N}_{2}\}\{\text{O}_{2}\}} = \frac{(0.034 \text{mol dm}^{-3})^{2}}{(0.17 \text{mol dm}^{-3})(0.17 \text{mol dm}^{-3})} = 4.0 \times 10^{-2} \]

*Q*is larger than*K*_{c}, so the reaction is not at equilibrium.- Because
*Q*is larger than*K*_{c}, the concentration of the product, NO, is larger than its equilibrium concentration and the concentrations of the reactants, N_{2}and O_{2}, are smaller than their equilibrium concentrations. Therefore some of the product, NO, will be consumed and more of the reactants, N_{2}and O_{2}, will be formed. - Use the given concentrations as the initial concentrations of reactants and product. Enter these into an ICE table. Let
*x*be the increase in the concentration of N_{2}as the system reacts to equilibrium. The ICE table looks like this:

N_{2} |
O_{2} |
NO | |
---|---|---|---|

Initial concentration/mol dm^{-3} |
0.17 | 0.17 | 0.034 |

Change in concentration/mol dm^{-3} |
x |
x |
-2x |

Equilibrium concentration/mol dm^{-3} |
0.17 + x |
0.17 + x |
0.034 - 2x |

Next, substitute the equilibrium concentrations into the *K*_{c} expression and solve for *x*.

\[K_{\text{c}} = 1.7 \times {10^{ - 3}} = \frac{(0.034 - 2x)^{2}}{(0.17 + x)(0.1 + x)}\]

Now take the square root of both sides of this equation. This gives

\(\sqrt {1.7 \times {10^{ - 3}}} = 0.0412 = \dfrac{0.034 - 2x}{0.17 + x}\)

Multiplying both sides by 0.17 + *x* gives

\( 0.0070 + 0.412 x = 0.034 - 2 x \)

\( 2 x + 0.0412 x =0.034 - 0.0070 \)

\(x = \dfrac{0.0270}{2.0412} = 0.0132\)

\( \text{[N}_{2} ] = \text{[O}_{2} ] = 0.17 + 0.013 = 0.183 \text{ mol dm}^{-3} \)

\( \text{[NO] } = 0.034 - 2(0.0132) = 0.0076 \text{ mol dm}^{-3} \)

Check the result by substituting these concentrations into the equilibrium constant expression.

\[K_{\text{c}} = \frac{(0.0076)^{2}}{(0.18)(0.18)} = 1.8 \times {10^{ - 3}}\]

This agrees to two significant figures with the *K*_{c} value of 1.7 x 10^{-3}.

## Contributors
Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.

Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.