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AP®ChemistryCollege BoardStudy GuidesUnit 7: EquilibriumEquilibrium ConstantsCalculating the Equilibrium Constant

Calculating the Equilibrium Constant (College Board AP® Chemistry)

Study Guide

Fallon

Written by: Fallon

Reviewed by: Stewart Hird

Calculating the Equilibrium Constant

  • For the generic equation:

aA + bB ⇌ cC + dD

Where a, b, c, and d represent stoichiometric coefficients and A, B, C, and D represent chemical species

  • The equilibrium expression for Kc is:

 begin mathsize 14px style straight K subscript straight c equals fraction numerator stretchy left square bracket straight C stretchy right square bracket to the power of straight c stretchy left square bracket straight D stretchy right square bracket to the power of straight d over denominator stretchy left square bracket straight A stretchy right square bracket to the power of straight a stretchy left square bracket straight B stretchy right square bracket to the power of straight b end fraction end style

           [A] indicates the concentration of species A in mol/L

  • The equilibrium expression KP is:

 begin mathsize 14px style K subscript p equals fraction numerator open parentheses P subscript C close parentheses to the power of c open parentheses P subscript D close parentheses to the power of d over denominator open parentheses P subscript A close parentheses to the power of a open parentheses P subscript B close parentheses to the power of b end fraction end style

           PA indicates the partial pressure of species A in atm

Worked Example

H2 (g) + CO2 (g) ⇌ H2O (g) + CO (g)

A mixture of H2, CO2, H2O, and CO gases is combined in a 1.0 L previously evacuated, rigid sealed container. The gases are allowed to react, according to the equation above, at a constant temperature of 700°C. A graph of the concentration of each gas as a function of time is shown below.

equilibrium-concentration-graph

What is the value of Kc for this reaction at 700°C?

Answer:

  • For the generic equation

aA + bB ⇌ cC + dD

The equilibrium expression for Kc is

 straight K subscript straight c equals fraction numerator stretchy left square bracket straight C stretchy right square bracket to the power of straight c stretchy left square bracket straight D stretchy right square bracket to the power of straight d over denominator stretchy left square bracket straight A stretchy right square bracket to the power of straight a stretchy left square bracket straight B stretchy right square bracket to the power of straight b end fraction

  • So, for the given reaction the equilibrium expression is

begin mathsize 14px style straight K subscript straight c equals fraction numerator stretchy left square bracket straight H subscript 2 straight O stretchy right square bracket stretchy left square bracket CO stretchy right square bracket over denominator stretchy left square bracket straight H subscript 2 stretchy right square bracket stretchy left square bracket CO subscript 2 stretchy right square bracket end fraction end style

  • To solve for Kc we substitute the concentration of each species at equilibrium

  • Equilibrium is established when the concentrations of each species remain constant

  • From the graph, the equilibrium concentrations are

[H2O] = 0.8 M

[CO] = 0.2 M

[H2] = 0.6 M

[CO2] = 0.5 M

  • Substituting the concentrations into the equation for Kc 

straight K subscript straight c equals space fraction numerator 0.8 cross times 0.2 over denominator 0.6 cross times 0.5 end fraction

begin mathsize 14px style straight K subscript straight c equals fraction numerator 0.16 over denominator 0.3 end fraction end style

Kc = 0.5

  • Note that Kc is a unitless value

Worked Example

2NO2 (g) ⇌ 2NO (g) + O2 (g)

A sample of pure NO2 gas in a sealed rigid container decomposes at a constant temperature of 1000 K and a constant pressure of 6.00 atm according to the equation above. At equilibrium, the vessel is found to contain 0.200 mol of NO2, 1.700 mol of NO, and 1.100 mol of O2. Calculate the equilibrium constant for partial pressure, KP, for this reaction.

Answer:

  • For the generic equation

aA + bB ⇌ cC + dD

The equilibrium expression for KP is

begin mathsize 14px style K subscript p equals fraction numerator open parentheses P subscript C close parentheses to the power of c open parentheses P subscript D close parentheses to the power of d over denominator open parentheses P subscript A close parentheses to the power of a open parentheses P subscript B close parentheses to the power of b end fraction end style

  • For the given reaction the equilibrium expression is

       K subscript p equals fraction numerator open parentheses P subscript N O end subscript close parentheses squared open parentheses P subscript O subscript 2 end subscript close parentheses over denominator open parentheses P subscript N O subscript 2 end subscript close parentheses squared end fraction

  • To solve for KP we substitute the partial pressure of each species at equilibrium

  • Determining the total number of moles of the gas mixture

begin mathsize 16px style straight n subscript total equals straight n subscript left parenthesis NO subscript 2 right parenthesis end subscript plus straight n subscript NO plus straight n subscript left parenthesis straight O subscript 2 right parenthesis end subscript end style

straight n subscript total equals 0.200 space mol space plus space 1.700 space mol space plus space 1.100 space mol

straight n subscript total equals 3.000 space mol

  • Determining the partial pressure for each gas using its mole fraction and total pressure

NO2 

begin mathsize 16px style straight P subscript NO subscript 2 end subscript equals straight n subscript NO subscript 2 end subscript over straight n subscript total cross times straight P subscript total end style

straight P subscript NO subscript 2 end subscript equals fraction numerator 0.200 space mol over denominator 3.000 space mol end fraction cross times 6.00 space atm

straight P subscript NO subscript 2 end subscript equals 0.400 space atm

NO

straight P subscript NO equals straight n subscript NO over straight n subscript total cross times straight P subscript total

straight P subscript NO equals fraction numerator 1.700 space mol over denominator 3.000 space mol space end fraction cross times 6.00 space atm

straight P subscript NO equals 3.40 space atm

O2

straight P subscript straight O subscript 2 end subscript equals straight n subscript straight O subscript 2 end subscript over straight n subscript total cross times straight P subscript total

straight P subscript straight O subscript 2 end subscript equals fraction numerator 1.100 space mol over denominator 3.000 space mol end fraction cross times 6.00 space atm

straight P subscript straight O subscript 2 end subscript equals 2.20 space atm

  • Substituting the partial pressures into the equation for KP 

begin mathsize 14px style straight K subscript straight P equals fraction numerator open parentheses 3.40 close parentheses squared space cross times 2.20 over denominator open parentheses 0.400 close parentheses squared end fraction end style

straight K subscript straight P equals fraction numerator 11.56 space cross times 2.20 over denominator 0.16 end fraction

KP=25.4320.16

KP=159

  • Note that KP is a unitless value

Examiner Tips and Tricks

When writing a Kc expression, always be sure to include the brackets, as brackets around a species represent the concentration of that species. However, when writing a KP expression never use brackets as this expression involves partial pressures not concentrations.

Make sure that the values substituted into a Kc expression have the unit mol/L and values substituted into a KP expression have the unit atm.

Remember that although the values substituted into a Kc or KP expression have units, the equilibrium constant is a unitless value.

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    Author
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    Fallon

    Author: Fallon

    Expertise: Chemistry Content Creator

    Fallon obtained a double major in chemistry and secondary education, and after graduating she taught Chemistry and Organic Chemistry for 7 years. Fallon’s passion for creating engaging classroom materials led her to pursue a career in content development. For over 3 years, Fallon has created videos, review materials, and practice questions for AP Chemistry, IGCSE, and other international exam boards.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.