Calculating Equilibrium Concentrations (College Board AP® Chemistry): Revision Note

A 2.0 M sample of cis-crotononitrile at 303°C is allowed to come to equilibrium according to the equation above. The equilibrium constant, K_c, for this reaction at 303°C is 1.40. What are the concentrations of cis-crotononitrile and trans-crotononitrile after equilibrium is established?

Answer:

Step 1: Completing an ICE table

• The initial concentrations of the cis- and trans- forms are 2.0 M and 0.0 M respectively

• In order to reach equilibrium, there must be an overall net conversion of reactants to products as no trans-crotononitrile is initially present

• Both stoichiometric coefficients are 1.

• We complete the ICE table as shown below

Step 2: Using an equilibrium expression to solve for x

• The equilibrium expression for K_c is

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{trans}-\mathrm{crotononitrile}\right]}{\left[\mathrm{cis}-\mathrm{crotononitrile}\right]}$

• Substituting the equilibrium constant and equilibrium concentrations into the equation and rearranging to solve for x

$1.40=\frac{\mathrm{x}}{2.00-\mathrm{x}}$

$1.40\times (2.00-\mathrm{x})=\mathrm{x}$

$2.80-1.40\mathrm{x}=\mathrm{x}$

$2.80=2.40\mathrm{x}$

$\frac{2.80}{2.40}=\mathrm{x}$

1.17 M = x

Step 3: Calculating the concentrations at equilibrium

• [cis-crotononitrile]_eq = 2.00 M – x

    = 2.00 M - 1.17 M

    = 0.83 M

• [trans-crotononitrile] = x

    = 1.17 M

I₂ (g) + H₂ (g) ⇌ 2HI (g)                    K_P = 160 at 500 K

Equimolar amounts of I₂ (g) and H₂ (g) were pumped into a previously evacuated, rigid sealed flask. The gases were allowed to react at 500 K according to the equation above. What is the partial pressure of HI (g) at equilibrium if the initial pressure inside the flask was 12.00 atm?

Answer:

Step 1: Completing an ICE table

• Initial partial pressure

  • ${\mathrm{P}}_{\mathrm{HI}\left(\mathrm{initial}\right)}=0\mathrm{atm}$ as no products are initially present

  • Equimolar amounts of I₂ and H₂ are initially present, so they must have equivalent mole fractions and therefore equal partial pressures

  • ${\mathrm{P}}_{{\mathrm{I}}_2\left(\mathrm{initial}\right)}={\mathrm{P}}_{{\mathrm{H}}_2\left(\mathrm{initial}\right)}=6.00\mathrm{atm}$ as the total initial pressure is 12.00 atm

• In order to reach equilibrium, there must be an overall net conversion of reactants to products as no HI is initially present

• The change in equilibrium must reflect the stoichiometric coefficients

• We complete the ICE table as shown below

Step 2: Using an equilibrium expression to solve for x

• The equilibrium expression for K_P is

${\mathrm{K}}_{\mathrm{P}}=\frac{{\left({\mathrm{P}}_{\mathrm{HI}}\right)}^2}{\left({\mathrm{P}}_{{\mathrm{I}}_2}\right)\left({\mathrm{P}}_{{\mathrm{H}}_2}\right)}$

• Substituting the equilibrium constant and partial pressures at equilibrium into the equation and rearranging to solve for x

$160=\frac{{\left(2\mathrm{x}\right)}^2}{\left(6.00-\mathrm{x}\right)\left(6.00-\mathrm{x}\right)}$

• Note that the right side of the equation is a perfect square

$\sqrt{160}=\sqrt{\frac{{\left(2\mathrm{x}\right)}^2}{(6.00-\mathrm{x})(6.00-\mathrm{x})}}$

$12.649=\frac{2\mathrm{x}}{6.00-\mathrm{x}}$

$12.649\times (6.00-\mathrm{x})=2\mathrm{x}$

$75.9-12.649\mathrm{x}=2\mathrm{x}$

$75.9=14.649\mathrm{x}$

$\frac{75.9}{14.649}=\mathrm{x}$

$5.18\mathrm{atm}=\mathrm{x}$

Step 3: Calculating the partial pressure of HI at equilibrium

• From the ICE table

${\mathrm{P}}_{\mathrm{HI}\left(\mathrm{eq}\right)}=2\mathrm{x}$

• Substituting x into the equation

${\mathrm{P}}_{\mathrm{HI}\left(\mathrm{eq}\right)}=2\times 5.18\mathrm{atm}$

${\mathrm{P}}_{\mathrm{HI}\left(\mathrm{eq}\right)}=10.4\mathrm{atm}$

Check your calculations by substituting your equilibrium concentrations or partial pressures back into the equilibrium expression. The value you determine for K should be close to the original value given in the question.

Always check the equilibrium expression for perfect squares. Taking the square root of both sides of the equation in these instances allows you to quickly simplify the calculation.