# 13.4 Equilibrium calculations

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By the end of this section, you will be able to:
• Write equations representing changes in concentration and pressure for chemical species in equilibrium systems
• Use algebra to perform various types of equilibrium calculations

We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.

Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.

On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:

$2{\text{NH}}_{3}\left(g\right)\stackrel{}{⇌}{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)$

If a sample of ammonia decomposes in a closed system and the concentration of N 2 increases by 0.11 M , the change in the N 2 concentration, Δ[N 2 ], the final concentration minus the initial concentration, is 0.11 M . The change is positive because the concentration of N 2 increases.

The change in the H 2 concentration, Δ[H 2 ], is also positive—the concentration of H 2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H 2 is three times the change in the concentration of N 2 because for each mole of N 2 produced, 3 moles of H 2 are produced.

$\text{Δ}\left[{\text{H}}_{2}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\left[{\text{N}}_{2}\right]$
$=3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(0.11\phantom{\rule{0.2em}{0ex}}M\right)=0.33\phantom{\rule{0.2em}{0ex}}M$

The change in concentration of NH 3 , Δ[NH 3 ], is twice that of Δ[N 2 ]; the equation indicates that 2 moles of NH 3 must decompose for each mole of N 2 formed. However, the change in the NH 3 concentration is negative because the concentration of ammonia decreases as it decomposes.

$\text{Δ}\left[{\text{NH}}_{3}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}-2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\left[{\text{N}}_{2}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}-2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(0.11\phantom{\rule{0.2em}{0ex}}M\right)=-0.22\phantom{\rule{0.2em}{0ex}}M$

We can relate these relationships directly to the coefficients in the equation

$\begin{array}{ccccc}\hfill 2{\text{NH}}_{3}\left(g\right)\hfill & \hfill \underset{}{\overset{}{⇌}}\hfill & \hfill {\text{N}}_{\text{2}}\left(g\right)\hfill & \hfill +\hfill & \hfill 3{\text{H}}_{2}\left(g\right)\hfill \\ \hfill \text{Δ}\left[{\text{NH}}_{3}\right]=-2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\left[{\text{N}}_{2}\right]\hfill & \hfill \hfill & \hfill \text{Δ}\left[{\text{N}}_{2}\right]=0.11\phantom{\rule{0.2em}{0ex}}M\hfill & \hfill \hfill & \hfill \text{Δ}\left[{\text{H}}_{2}\right]=3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\left[{\text{N}}_{2}\right]\hfill \end{array}$

Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.

If we did not know the magnitude of the change in the concentration of N 2 , we could represent it by the symbol x .

$\text{Δ}\left[{\text{N}}_{2}\right]=x$

The changes in the other concentrations would then be represented as:

$\text{Δ}\left[{\text{H}}_{2}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\left[{\text{N}}_{2}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}3x$
$\text{Δ}\left[{\text{NH}}_{3}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}-2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}\left[{\text{N}}_{2}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}-2x$

The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.

$\begin{array}{ccccc}2{\text{NH}}_{3}\left(g\right)\hfill & \underset{}{\overset{}{⇌}}\hfill & {\text{N}}_{2}\left(g\right)\hfill & +\hfill & 3{\text{H}}_{2}\left(g\right)\hfill \\ -2x\hfill & \hfill & x\hfill & \hfill & 3x\hfill \end{array}$

The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.

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