Representations of Equilibrium (College Board AP® Chemistry)

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Fallon

Written by: Fallon

Reviewed by: Stewart Hird

Representations of Equilibrium

  • Particulate diagrams can be used to represent a reversible reaction over time

  • When no observable changes occur to the number of particles in a system involving a reversible reaction, the reaction has reached a state of equilibrium

  • Consider the following reaction over a period of 80 seconds

N2O4 (g) ⇌ 2NO2 (g)

Particulate representation for a reversible reaction

Particulate representation for the reversible reaction N2O4 (g) ⇌ 2NO2 (g) showing that equilibrium is established when the number of particles of each species remains constant over time

  • A particulate diagram of a reversible reaction at a certain point in time can be used to calculate K

  • For the generic equation:

aA + bB ⇌ cC + dD

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

  • A particulate representation can also be used to calculate Q and determine if the reversible reaction has reached equilibrium

    • The equations for Qc and QP are identical to those for K, but Q can be determined using concentration or partial pressure values from any point during the reaction

    • At equilibrium, Q = K

Worked Example

A2 (g) + 2B (g) ⇌ A2B2 (g)

Gases A2 and B are placed in an empty 1.00 L rigid vessel at a constant temperature of 25°C. The particle diagram below represents the contents of the vessel after equilibrium is established. Assume that each particle in the diagram represents 1 mole. Based on this information, what is the equilibrium constant, Kc, for the reaction above at 25°C?

particle-representation-at-equilibrium

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 the equilibrium expression for Kc for the given reaction is

 straight K subscript straight c equals fraction numerator stretchy left square bracket straight A subscript 2 straight B subscript 2 stretchy right square bracket over denominator stretchy left square bracket straight A subscript 2 stretchy right square bracket stretchy left square bracket straight B stretchy right square bracket squared end fraction

  • Each particle in the diagram represents 1.00 mole and the total volume of the container is 1.00 L

  • Therefore, the number of particles of a species in the diagram also represents the concentration of that species at equilibrium in mol/L

  • The equilibrium concentrations are

    • [A2B2]eq = 4.00 mol/L

    • [A2]eq = 2.00 mol/L

    • [B]eq = 4.00 mol/L

  • Substituting the concentrations into the equilibrium expression

straight K subscript straight c equals fraction numerator 4.00 over denominator 2.00 cross times open parentheses 4.00 close parentheses squared end fraction

straight K subscript straight c equals fraction numerator 4.00 over denominator 2.00 space cross times 16.00 end fraction

straight K subscript straight C equals fraction numerator 4.00 over denominator 32.00 end fraction

straight K subscript straight C equals 0.125

Worked Example

2X (g) ⇌ Y (g)

At 25°C, the equilibrium constant, Kc, for the reaction shown above is 15. Which of the following particle diagrams best represents the system at equilibrium? Assume that each particle in the diagrams represents a concentration of 0.1 M.

particle-representation-q-versus-k

Answer:

Step 1: Determine the expression for Q

  • For the generic equation

aA + bB ⇌ cC + dD

The equilibrium expression for Qc is

straight Q subscript straight C equals fraction numerator open square brackets straight C close square brackets to the power of straight c open square brackets straight D close square brackets to the power of straight d over denominator open square brackets straight A close square brackets to the power of straight a open square brackets straight B close square brackets to the power of straight b end fraction

  • So the expression for Qc for the given reaction is

straight Q subscript straight C equals fraction numerator open square brackets straight Y close square brackets over denominator open square brackets straight X close square brackets squared end fraction

Step 2: Solve for Q

  • Diagram A

    • [Y] = 0.4 M

    • [X] = 0.4 M

straight Q subscript straight C equals fraction numerator 0.4 over denominator 0.4 squared end fraction

straight Q subscript straight C equals fraction numerator 0.4 over denominator 0.16 end fraction

straight Q subscript straight C equals 2.5

  • Diagram B

    • [Y] = 0.3 M

    • [X] = 0.5 M

straight Q subscript straight C equals fraction numerator 0.3 over denominator 0.5 squared end fraction

straight Q subscript straight C equals fraction numerator 0.3 over denominator 0.25 end fraction

straight Q subscript straight C equals 1.2

  • Diagram C

    • [Y] = 0.6 M

    • [X] = 0.2 M

straight Q subscript straight C equals fraction numerator 0.6 over denominator 0.2 squared end fraction

straight Q subscript straight C equals fraction numerator 0.6 over denominator 0.04 end fraction

straight Q subscript straight C equals 15

  • Diagram D

    • [Y] = 0.5 M

    • [X] = 0.3 M

straight Q subscript straight C equals fraction numerator 0.5 over denominator 0.3 squared end fraction 

straight Q subscript straight C equals fraction numerator 0.5 over denominator 0.09 end fraction

straight Q subscript straight C equals 5.6

  • A system is in equilibrium when Q = K

  • So diagram C best represents the system at equilibrium

Examiner Tips and Tricks

Be sure to read a particulate diagram question carefully. An individual particle may represent one or more moles of the substance or a specific concentration. In addition, the size of the vessel, while often 1 L, may vary.

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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.