#### Learning Objective

- Calculate the concentrations of reaction components at equilibrium given the starting concentrations and the equilibrium constant

#### Key Points

- If you know K
_{c}and the initial concentrations for a reaction, you can calculate the equilibrium concentrations. - Using the ICE chart and equilibrium-constant equation, you can write an expression to describe the concentration changes in the reactants and products.

#### Terms

- concentrationThe proportion of a substance in a mixture.
- equilibriumThe state of a reaction in which the rates of the forward and reverse reactions are the same.
- reaction rateHow fast or slowly a reaction takes place.

## The ICE Chart

The theory of chemical equilibrium tells us that the species involved in a reversible reaction will eventually arrive at constant concentrations. Let’s consider the following reaction:

[latex]N_2 + O_2 \rightleftharpoons 2NO[/latex]

Initially, the concentration for each species is as follows: [N_{2}]_{0} = [O_{2}]_{0} = 0.1 M, and [NO]_{0} = 0 M. The value of Kc for this reaction is known to be 0.1 (we will assume that this experimentally determined quantity is given to us). We can write the following K_{c} expression based on this starting information:

[latex]K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}=0.1[/latex]

Since we know what our K_{c} value is and the initial concentrations of reactants, we can set up an ICE chart to track the changes in concentration, as the reaction proceeds towards equilibrium. ICE stands for “initial, change, equilibrium.”

Note that the values of *x* in the third row of our chart indicate the change in concentration of each species over the course of the reaction. The coefficients present in the balanced equation tell us how many moles of atoms or molecules participate in the reaction. Using these coefficients, the balanced equation tells us that for every mole of N_{2} and mole O_{2} consumed, 2 moles of NO are produced. We can designate *x* as the change in concentration of N_{2} and O_{2. } As reactants located on the left hand side of our balanced equation, the sign is negative as they are being consumed. Similarly, we designate +2*x* as the change in concentration for NO, but it’s positive because it’s being produced. After we fill in our chart we can determine the equilibrium concentrations by adding down the columns of the ICE chart. The equilibrium concentrations for each species are therefore: [N_{2}] = 0.1 – x; [O_{2}] = 0.1 – x; [NO] = 2x.

## Plugging into the K_{C} Expression and Solving for *x*

Now that we have expressions for the equilibrium concentrations of each species, we can substitute them into our expression for K_{c}:

[latex]K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}=\frac{(2x)^{2}}{(0.1 - x)(0.1 - x)}[/latex]

If we expand, collect terms, and solve for *x*, we get the following quadratic equation:

[latex]x\quad =\quad \frac {{-0.2}{ { { K }_{ c } }{ \pm }{ \sqrt { \left( .2{ K }_{ c } \right) ^{ 2 }{ -4\left( 4-{ K }_{ c } \right) \left( -0.01{ K }_{ c } \right) } } } } }{ { 2 }\left( 4-{ K }_{ c } \right) }[/latex]

When solving a quadratic equation, we will always get two values for *x. *The two x values are -0.1188M and 0.0137M. Only one of these values involves equilibrium concentrations that are actually possible. We can determine which x value is the real solution by substituting it into our equilibrium concentrations, found on the ICE chart. For example, consider the value x = −0.0188. Substituting this into the equilibrium amount for N_{2} gives a concentration of 0.1 – (-0.0188) = 0.1188 M. This is clearly impossible, since we cannot have more N_{2} at equilibrium than we had at the beginning. Therefore, we use the other root for *x*,which is 0.0137.

Knowing the initial concentration values and equilibrium constant we were able to calculate the equilibrium concentrations for N_{2}, O_{2} and NO. In the system we evaluated, at equilibrium we would expect to find that [O_{2}]_{eq} = [N_{2}]_{eq} = 0.086 M and [NO]_{eq} = 0.028 M. Note that we could have solved for the amount of NO produced rather than for the amount of N_{2} and O_{2} consumed. The result would be the same.