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What does the value of the equilibrium constant (K) tell us about a reaction at chemical equilibrium?

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Equilibrium constants are always calculated using the product and reactant concentrations at equilibrium. Study the examples on each tab to learn more about the value of the equilibrium constant.

Look at this graph of concentrations over a period of time. At what point has this reaction reached equilibrium?

Full image description in following section.

Graph of the product and reactant of a reaction. The y-axis is labelled “Concentration,” and the x-axis is labeled “Time.” The graph of the product curves upward from the left to the right, then the curve flattens out. The graph of the reactant curves downward from the left to the right, then the curve flattens out.

Full image description in following section.

Graph of the product and reactant of a reaction. The y-axis is labelled “Concentration,” and the x-axis is labeled “Time.” The graph of the product curves upward from the left to the right, then the curve flattens out. The graph of the reactant curves downward from the left to the right, then the curve flattens out. There is a vertical dashed green line toward the right of the graph where each curve flattens out.

Remember that the concentrations of the reactants and products remain constant once a reaction has reached equilibrium. On the graph, this is shown when the slope of each curve flattens out to zero (i.e., a horizontal line). This point in time is represented by the dashed green line on the graph shown here.

The value of the equilibrium constant (K) tells what the reaction looks like at equilibrium--that is, whether there are more products or reactants present. Because the ratio used to determine the equilibrium constant has the products as the numerator, a large concentration of products will yield a large K value. Conversely, a large concentration of reactants (which is the denominator of the equilibrium constant ratio) will yield a small K value.

\(K = \ \frac{\lbrack \text{products}\rbrack}{\lbrack \text{reactants}\rbrack}\)

In the graph above, there are more products at equilibrium. Will this reaction have a large or small value for K?

This reaction will have a large value for K because the concentration of the products is much greater than the concentration of the reactants at equilibrium.

The value of the equilibrium constant gives us important information about what the reaction looks like at equilibrium. A very large K value means there are mostly products present at equilibrium, while a very small K value means there are mostly reactants present at equilibrium. For example:

Sulfur dioxide gas is a compound that has many uses, including its use as a disinfectant. Sulfur dioxide reacts with oxygen to produce sulfur trioxide, according to the equation below.

2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)

Cleaner's hand in blue rubber protective glove holding a white chemical spray bottle.

The equilibrium constant expression is a ratio of the concentration of the products over the concentration of the reactants (with coefficients becoming exponents). For the reaction described above, \(K = \ \frac{\lbrack\text{SO}_{3}\rbrack^{2}}{\lbrack\text{SO}_{2}\rbrack^{2}\lbrack \text{O}_{2}\rbrack}\). And the equilibrium constant (K) = 2.21 x 1012. What does it mean for a reaction to have such a large value for K?

Because the equilibrium constant is a very large number, this means there are more products (SO3) present at equilibrium than reactants.

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Let's Practice

Complete this activity to practice using equilibrium constants to describe the equilibrium position.

What does equilibrium look like?

For the reaction HF (aq) ⇌ H+ (aq) + F- (aq), K = 6.6 x 10-4

more reactants present
more products present
equal amounts of reactants and products

For the reaction HCl (aq) ⇌ H+ (aq) + Cl- (aq), K = 1.3 x 106

more reactants present
more products present
equal amounts of reactants and products

For the reaction NaHCO3 (aq) ⇌ CO2 (g) + NaOH (aq), K = 1.83 x 10-14

more reactants present
more products present
equal amounts of reactants and products

For the reaction 2H2 (g) + O2 (g) ⇌ 2H2O (g), K = 2.4 x 1047

more reactants present
more products present
equal amounts of reactants and products

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