Chemistry » Fundamental Equilibrium Concepts » Equilibrium Calculations

Equilibrium Calculations

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 N2 increases by 0.11 M, the change in the N2 concentration, Δ[N2], the final concentration minus the initial concentration, is 0.11 M. The change is positive because the concentration of N2 increases.

The change in the H2 concentration, Δ[H2], is also positive—the concentration of H2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H2 is three times the change in the concentration of N2 because for each mole of N2 produced, 3 moles of H2 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 NH3, Δ[NH3], is twice that of Δ[N2]; the equation indicates that 2 moles of NH3 must decompose for each mole of N2 formed. However, the change in the NH3 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 N2, 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.

Example

Determining Relative Changes in Concentration

Complete the changes in concentrations for each of the following reactions.

(a) \(\begin{array}{lcccc} \text{C}_2\text{H}_2(g) & + & 2\text{Br}_2(g) & {\rightleftharpoons} & \text{C}_2\text{H}_2\text{Br}_4(g) \\[0.5em] x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}\)

(b) \(\begin{array}{lcccc} \text{I}_2(aq) & + & \text{I}^{-}(aq) & {\rightleftharpoons} & \text{I}_3^{\;\;-}(aq) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & x \end{array}\)

(c) \(\begin{array}{lcccccc} \text{C}_3\text{H}_8(g) & + & 5\text{O}_2(g) & {\rightleftharpoons} & 3\text{CO}_2(g) & + & 4\text{H}_2\text{O}(g) \\[0.5em] x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}\)

Solution

(a) \(\begin{array}{cccc}{\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\hfill & 2{\text{Br}}_{2}\left(g\right)\hfill & ⇌\hfill & {\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}\left(g\right)\hfill \\ x\hfill & 2x\hfill & & -x\hfill \end{array}\)

(b) \(\begin{array}{cccc}{\text{I}}_{2}\left(aq\right)+\hfill & {\text{I}}^{\text{−}}\left(aq\right)\hfill & ⇌\hfill & {\text{I}}_{3}{}^{\text{−}}\left(aq\right)\hfill \\ -x\hfill & -x\hfill & & x\hfill \end{array}\)

(c) \(\begin{array}{lllll}{\text{C}}_{3}{\text{H}}_{8}\left(g\right)+\hfill & 5{\text{O}}_{2}\left(g\right)\hfill & ⇌\hfill & 3{\text{CO}}_{2}\left(g\right)+\hfill & 4{\text{H}}_{2}\text{O}\left(g\right)\hfill \\ x\hfill & 5x\hfill & & -3x\hfill & -4x\hfill \end{array}\)

[Attributions and Licenses]


This is a lesson from the tutorial, Fundamental Equilibrium Concepts and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts