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Calculating the Equilibrium Constant

Caleb Thomas

Caleb Thomas

9 min read

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Study Guide Overview

This study guide covers calculating equilibrium constants (Kc and Kp). It explains the formulas, provides example calculations, and discusses the meaning of K values. The guide also includes tips for calculations, a sample AP question, and practice problems with an answer key. Key concepts include equilibrium, stoichiometric coefficients, partial pressures, and molar concentrations.

Chemical Equilibrium: Calculating Equilibrium Constants

Introduction

Now that you're familiar with the concepts of KcK_c and KpK_p, let's dive into how to calculate them! This guide will help you master these calculations, ensuring you're ready for anything the AP Chemistry exam throws your way. Remember, equilibrium is all about the balance between reactants and products. Let's get started!

Equilibrium calculations are a cornerstone of AP Chemistry. Expect to see them in both multiple-choice and free-response questions. Mastering this topic is crucial for a strong exam performance.

Equilibrium Constant Formulas

Key Concepts

  • Equilibrium: A state where the rate of forward and reverse reactions are equal, and the concentrations of reactants and products remain constant.

  • KcK_c (Equilibrium Constant in terms of concentration): Relates the concentrations of reactants and products at equilibrium.

  • KpK_p (Equilibrium Constant in terms of pressure): Relates the partial pressures of gaseous reactants and products at equilibrium.

Key Concept

Remember, solids and pure liquids are NOT included in the equilibrium expression. Only aqueous solutions and gases are considered.

Formulas

Both KcK_c and KpK_p follow the same basic formula: products over reactants, each raised to the power of their stoichiometric coefficients. The difference lies in whether you're using concentrations (for KcK_c) or partial pressures (for KpK_p).

KcK_c Formula

latex
<math-block>K\_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}</math-block>

Where:

  • [A], [B] are the equilibrium concentrations of reactants.
  • [C], [D] are the equilibrium concentrations of products.
  • a, b, c, d are the stoichiometric coefficients from the balanced chemical equation.

KpK_p Formula

latex
<math-block>K\_p = \frac{P\_C^cP\_D^d}{P\_A^aP\_B^b}</math-block>

Where:

  • PAP_A, PBP_B are the partial pressures of reactants at equilibrium.

  • PCP_C, PDP_D are the partial pressures of products at equilibrium.

  • a, b, c, d are the stoichiometric coefficients from the balanced chemical equation.

Exam Tip

Pay close attention to the units! KcK_c uses molar concentrations (mol/L), while KpK_p uses partial pressures (usually in atm).

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Image Courtesy of ScienceAid

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Image Courtesy of Quora

Calculating KcK_c

Example

Let's calculate KcK_c for the following reaction:

latex
<math-block>CO\_2(g) + H\_2(g) \rightleftharpoons CO(g) + H\_2O(g)</math-block>

Given:

  • 0.1908 moles of CO2CO_2
  • 0.0908 moles of H2H_2
  • 0.0092 moles of COCO
  • 0.0092 moles of H2OH_2O
  • Volume = 2.00 L

Step-by-Step Solution

  1. Write the KcK_c expression:

latex Kc=[CO][H2O][CO2][H2]K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} ```

  1. Calculate equilibrium concentrations:

    • [CO]=0.0092 mol/2.00 L=0.0046 M[CO] = 0.0092 \text{ mol} / 2.00 \text{ L} = 0.0046 \text{ M}
    • [H2O]=0.0092 mol/2.00 L=0.0046 M[H_2O] = 0.0092 \text{ mol} / 2.00 \text{ L} = 0.0046 \text{ M}
    • [CO2]=0.1908 mol/2.00 L=0.0954 M[CO_2] = 0.1908 \text{ mol} / 2.00 \text{ L} = 0.0954 \text{ M}
    • [H2]=0.0908 mol/2.00 L=0.0454 M[H_2] = 0.0908 \text{ mol} / 2.00 \text{ L} = 0.0454 \text{ M}
  2. Plug the concentrations into the KcK_c expression:

latex Kc=(0.0046)(0.0046)(0.0954)(0.0454)=4.9×103K_c = \frac{(0.0046)(0.0046)}{(0.0954)(0.0454)} = 4.9 \times 10^{-3} ```

Quick Fact

KcK_c is unitless, but always make sure you have the correct units for your concentrations (mol/L).

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Image Courtesy of Purdue University

Calculating KpK_p

Example

Let's calculate KpK_p for the following reaction:

latex
<math-block>2N\_2O\_5(g) \rightleftharpoons O\_2(g) + 4NO\_2(g)</math-block>

Given:

  • P(N2O5)=2.00 atmP(N_2O_5) = 2.00 \text{ atm}
  • P(O2)=0.296 atmP(O_2) = 0.296 \text{ atm}
  • P(NO2)=1.70 atmP(NO_2) = 1.70 \text{ atm}

Step-by-Step Solution

  1. Write the KpK_p expression:

latex Kp=PO2PNO24PN2O52K_p = \frac{P_{O_2}P_{NO_2}^4}{P_{N_2O_5}^2} ```

  1. Plug in the partial pressures:

latex Kp=(0.296)(1.70)4(2.00)2=0.618K_p = \frac{(0.296)(1.70)^4}{(2.00)^2} = 0.618 ```

Understanding the Equilibrium Constant

What Does K Tell Us?

The equilibrium constant (K) is a ratio that tells us the extent to which a reaction will proceed. It compares the amount of products to the amount of reactants at equilibrium.

  • K > 1: Product-favored reaction. At equilibrium, there are more products than reactants.

  • K < 1: Reactant-favored reaction. At equilibrium, there are more reactants than products.

  • K = 1: The amount of reactants and products are roughly equal at equilibrium.

Memory Aid

Think of K as a seesaw: if K is greater than 1, the seesaw tips toward the products; if K is less than 1, it tips toward the reactants. ⚖️

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Image Courtesy of Labster Theory

Tips for Calculating Equilibrium Constants

  • Equilibrium Values Only: Make sure you're using concentrations or partial pressures at equilibrium. Using values at any other point will calculate the reaction quotient (Q), not K.

  • Correct Units: Always use molarity (mol/L) for KcK_c and partial pressures (usually atm) for KpK_p.

  • Unit Conversions: Be prepared to convert grams to moles, moles to molarity, or total pressure to partial pressures using the ideal gas law and mole fractions.

  • Stoichiometry: Don't forget to raise the concentrations/pressures to the power of their stoichiometric coefficients in the balanced equation.

Common Mistake

A very common mistake is to use initial concentrations instead of equilibrium concentrations. Always double-check that the values you're using are at equilibrium!

AP Question - 2017 #3

Let's look at a real AP question to see how these concepts are applied.

latex
<math-block>N\_2(g) + O\_2(g) \rightleftharpoons 2NO(g)</math-block>

At high temperatures, N2(g)N_2(g) and O2(g)O_2(g) react to produce nitrogen monoxide, NO(g)NO(g), as represented by the equation above.

(a) Write the expression for the equilibrium constant, KpK_p, for the forward reaction.

Solution

latex
<math-block>K\_p = \frac{P\_{NO}^2}{P\_{N\_2}P\_{O\_2}}</math-block>
Exam Tip

In free-response questions, always show your work, including the equilibrium expression before plugging in values. This can earn you partial credit even if your final answer is incorrect.

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Final Exam Focus

High-Priority Topics

  • Writing KcK_c and KpK_p expressions: Make sure you know how to write the correct expressions, including the stoichiometric coefficients.
  • Calculating equilibrium constants: Practice calculating KcK_c and KpK_p from given equilibrium concentrations or partial pressures.
  • Understanding the meaning of K: Know what a K value greater than, less than, or equal to 1 means for the reaction.
  • Unit conversions: Be comfortable converting between grams, moles, molarity, and partial pressures.

Common Question Types

  • Multiple Choice: Expect questions that test your understanding of the concepts and your ability to write the correct equilibrium expressions.
  • Free Response: Look for questions that require you to calculate KcK_c or KpK_p from given data, often involving unit conversions and multiple steps.

Last-Minute Tips

  • Double-check your units: Always make sure you're using the correct units for concentrations and pressures.

  • Show your work: Even if you make a mistake, you can still earn partial credit for showing your steps.

  • Manage your time: Don't spend too much time on any one question. If you're stuck, move on and come back to it later.

  • Stay calm: Take a deep breath and trust your preparation. You've got this!

Practice Question

Practice Questions

Multiple Choice Questions

  1. For the reaction 2SO2(g)+O2(g)2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), the equilibrium constant KpK_p is 4.0 atm⁻¹. If the partial pressures of SO2SO_2 and O2O_2 at equilibrium are 0.20 atm and 0.10 atm respectively, what is the partial pressure of SO3SO_3 at equilibrium?

    (A) 0.04 atm (B) 0.08 atm (C) 0.16 atm (D) 0.40 atm

  2. The equilibrium constant, KcK_c, for the reaction H2(g)+I2(g)2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g) is 50 at 400°C. If 2.0 moles of H2H_2 and 2.0 moles of I2I_2 are placed in a 1.0 L container, what is the equilibrium concentration of HI?

    (A) 0.22 M (B) 1.56 M (C) 3.11 M (D) 3.50 M

Free Response Question

Consider the following reaction:

latex
<math-block>N\_2(g) + 3H\_2(g) \rightleftharpoons 2NH\_3(g)</math-block>

Initially, a 10.0 L vessel contains 1.00 mol of N2N_2 and 3.00 mol of H2H_2 at 500 K. At equilibrium, the concentration of NH3NH_3 is 0.060 M.

(a) Calculate the initial concentrations of N2N_2 and H2H_2.

(b) Calculate the equilibrium concentrations of N2N_2 and H2H_2.

(c) Calculate the value of the equilibrium constant, KcK_c, for the reaction.

(d) Calculate the value of KpK_p for the reaction.

Answer Key

Multiple Choice Questions

  1. (D) 0.40 atm
  2. (C) 3.11 M

Free Response Question

(a) Initial concentrations:

  • [N2]=1.00 mol/10.0 L=0.100 M[N_2] = 1.00 \text{ mol} / 10.0 \text{ L} = 0.100 \text{ M} (1 point)
  • [H2]=3.00 mol/10.0 L=0.300 M[H_2] = 3.00 \text{ mol} / 10.0 \text{ L} = 0.300 \text{ M} (1 point)

(b) Equilibrium concentrations:

  • Change in [NH3]=0.060 M[NH_3] = 0.060 \text{ M} (given). From the balanced equation, the change in [N2][N_2] is -0.030 M and the change in [H2][H_2] is -0.090 M.
  • [N2]=0.1000.030=0.070 M[N_2] = 0.100 - 0.030 = 0.070 \text{ M} (1 point)
  • [H2]=0.3000.090=0.210 M[H_2] = 0.300 - 0.090 = 0.210 \text{ M} (1 point)

(c) KcK_c calculation:

latex Kc=[NH3]2[N2][H2]3=(0.060)2(0.070)(0.210)3=11.7K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.060)^2}{(0.070)(0.210)^3} = 11.7 ``` (2 points for correct setup and answer)

(d) KpK_p calculation:

latex Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n} ```

  • Δn=2(1+3)=2\Delta n = 2 - (1 + 3) = -2 (1 point)

latex Kp=11.7(0.0821×500)2=0.0069K_p = 11.7(0.0821 \times 500)^{-2} = 0.0069 ``` (1 point)

Question 1 of 10

Given the reaction: N2(g)+3H2(g)2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), which of the following is the correct KcK_c expression? 🎉

Kc=[N2][H2]3[NH3]2K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}

Kc=[NH3]2[N2][H2]3K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

Kc=[NH3][N2][H2]K_c = \frac{[NH_3]}{[N_2][H_2]}

Kc=[N2][H2][NH3]K_c = \frac{[N_2][H_2]}{[NH_3]}