Equilibrium Calculations - from Initial Concentrations

Another common type of equilibrium calculation is one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:

Identify the direction in which the reaction will proceed to reach equilibrium.
(Compare Q and K)
Build an ICE table.
Calculate the concentration changes and equilibrium concentrations.
Confirm the calculated equilibrium concentrations.

The next two example exercises demonstrate the application of this strategy.

Calculation of Equilibrium Concentrations

Under certain conditions, the equilibrium constant $\mathit K_c$ for the decomposition of $PCl_5\mathit (g)$ into $PCl_3\mathit (g)$ and $Cl_2\, (g)$ is 0.0211. What are the equilibrium concentrations of $PCl_5$, $PCl_3$, and $Cl_2$ in a mixture that initially contained only $PCl_5$ at a concentration of 1.00 $\mathit M$

Solution

Use the stepwise process described earlier.

Identify the direction in which the reaction will proceed to reach equilibrium.

The balanced equation for the decomposition of $PCl_5$ is
$$PCl_5\;(g) ⇌ PCl_3\;(g) + Cl_2\;(g)$$
Because only the reactant is present initially $\mathit {Q_c} = 0$ and the reaction will proceed to the right.
Develop an ICE table.
Solve for the change and the equilibrium concentrations.

Substituting the equilibrium concentrations into the equilibrium constant equation gives
$$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5}\\ 0.0211=\frac{(x)(x)}{(1.00−x)}\\ 0.0211=\frac{(x)(x)}{(1.00−x)}\\ 0.0211(1.00−x)=x^2\\ x^2+0.0211x-0.0211=0$$
We can use the quadratic equation to solve for $x$: see Appendix B for a refresher.
$$x=\frac{−b± \sqrt{b^2−4ac}}{2a}$$
In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:
$$x=\frac{−0.0211± \sqrt{(0.0211)^2−4(1)(-0.0211)}}{(2)(1)}\\ x=\frac{−0.0211± \sqrt{(4.45×10^{-4})+(8.44×10^{-2})}}{2}\\ x=\frac{−0.0211±0.291}{2}$$ The two roots of the quadratic are: $$x=\frac{−0.0211+0.291}{2}=0.135$$ and $$x=\frac{−0.0211-0.291}{2}=−0.156$$ For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so $\mathbf{x = 0.135\, \text{M}}$.

The equilibrium concentrations are
$$\mathbf{[PCl_5]=1.00-x=1.00−0.135=0.87\, \text{M}\\ [PCl_3]=x=0.135\, \text{M}\\ [Cl_2]=x=0.135\, \text{M}}$$
Confirm the calculated equilibrium concentrations.

Substitution into the expression for $K_c$ (to check the calculation) gives $$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}\\ K_C=\frac{(0.135)(0.135)}{0.87}\\ K_C=0.021$$ The equilibrium constant calculated from the equilibrium concentrations is equal to the value of $K_c$ given in the problem (when rounded to the proper number of significant figures).

Check Your Learning

Acetic acid, $CH_3CO_2H$, reacts with ethanol, $C_2H_5OH$, to form water and ethyl acetate, $CH_3CO_2C_2H_5$.

$$CH_3CO_2H+C_2H_5OH \rightleftharpoons CH_3CO_2C_2H_5+H_2O$$

The equilibrium constant $K_C$ for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH₃CO₂H, 0.15 M in C₂H₅OH, 0.40 M in CH₃CO₂C₂H₅, and 0.40 M in H₂O?

Answer

Following the steps described above:

Identify the direction in which the reaction will proceed to reach equilibrium.

We can identify the direction of the reaction by calculating the reaction quotient Q: $$Q = \frac{[CH_3 CO_2 C_2 H_5][H_2 O]}{[CH_3 CO_2 H][C_2 H_5 OH]}\\ Q = \frac{[0.40\, \text{M}][0.40\, \text{M}]}{[0.15\, \text{M}][0.15\, \text{M}]}\\ Q = 7.111\\ Q \gt K: \text{the reaction will proceed to the left (towards the reactants)}$$ (Since the solvent is specified as dioxane, the water is not a solvent / pure liquid in this reaction and will appear in the equilibrium expression).

Develop an ICE table.

	$CH_3CO_2H$	$+$	$C_2H_5OH $	$\rightleftharpoons $	$CH_3CO_2C_2H_5$	$+$	$H_2O$
Initial concentration	0.15 M		0.15 M		0.40 M		0.40 M
Change in concentration	+x		+x		-x		-x
Equilibrium concentration	0.15 + x		0.15 + x		0.40 – x		0.40 – x

Calculate the concentration changes and the equilibrium concentrations.

Using the expression for K: $$ K = \frac{[CH_3 CO_2 C_2 H_5][H_2 O]}{[CH_3 CO_2 H][C_2 H_5 OH]}\\ 4.0 = \frac{[0.15+x][0.15+x]}{[0.40-x][0.40-x]}\\ 4.0 = \frac{[0.15+x]^2}{[0.40-x]^2}\\ \sqrt{4.0} = \sqrt{\frac{[0.15+x]^2}{[0.40-x]^2}}\\ \pm 2.0 = \frac{[0.15+x]}{[0.40-x]}\\ 0.80-2.0x = 0.15+x \;\&\; -0.80+2.0x = 0.15+x\\ x = 0.21_{667} \;\&\; 0.95$$ Only $x = 0.21_{667}$ makes chemical sense (the other value would result in a negative concentration if used). $$ [CH_3CO_2H] = [C_2H_5OH] = 0.15 + x = 0.36_{6667} = \mathbf{0.37\, \text{M}} \\ [CH_3CO_2C_2H_5] = [H_2 O] = 0.40 – x = 0.18_{3333} = \mathbf{0.18\, \text{M}}$$
Confirm the calculated equilibrium concentrations.

Sanity check: $$K = \frac{[CH_3 CO_2 C_2 H_5][H_2 O]}{[CH_3 CO_2 H][C_2 H_5 OH]}\\ K = \frac{[0.36_{6667}\, \text{M}][0.36_{6667}\, \text{M}]}{[0.18_{3333}\, \text{M}][0.18_{3333}\, \text{M}]}\\ K = 4.0$$ The recalculated K agrees with the provided (actual) K – our calculations are correct.

Check Your Learning

A 1.00-L flask is filled with 1.00 mole of H₂ and 2.00 moles of I₂. The value of the equilibrium constant $K_C$ for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H₂, I₂, and HI in moles/L?

$$H_2\;(g)+I_2\;(g)\rightleftharpoons 2\,HI\;(g)$$

Answer

Following the steps described above:

No HI is present in this reaction ([HI] = 0). The reaction must proceed to the right (towards the products) in order to reach equilibrium.

	$H_2$	$+$	$I_2 $	$\rightleftharpoons $	$2\, HI$
Initial concentration	1.00 M		2.00 M		0 M
Change in concentration	-x		-x		+2x
Equilibrium concentration	1.00 – x		2.00 – x		2x

Using the expression for K: $$ K = \frac{[HI]^2}{[H_2][I_2]}\\ 50.5 = \frac{[2x]^2}{[1.00-x][2.00-x]}\\ 46.5x^2 – 151.5x + 101 = 0 \\ \text{Solve using the quadratic formula:}\\ x = 0.934_{984} \;\&\; 2.32_{308}$$ Only $x = 0.934_{984}$ makes chemical sense (the other value would result in a negative concentration if used). $$[H_2] = 1.00 – x = 0.06_{5016} = \mathbf{0.07\, \text{M}}\\ [I_2] = 2.00 – x = 1.06_{5016} = \mathbf{1.07\, \text{M}}\\ [HI] = 2x = 1.86_{9968} = \mathbf{1.87\, \text{M}}\\$$
Sanity check: $$K = \frac{[HI]^2}{[H_2][I_2]}\\ K = \frac{[1.86_{9968}]^2}{[0.06_{5016}][1.06_{5016}]}\\ K = 50.5$$ The recalculated K agrees with the provided (actual) K – our calculations are correct.