Your Dashboard username@email.com

⬅ Previous Lesson Workshop Index Next Lesson ➡

Reactions and Reactivity: Le Châtelier’s Principle and Equilibrium Constants

Objective

In this lesson we will use Le Châtelier’s principle to determine changes in concentration that will occur at equilibrium. We will also look at equilibrium constants of reactions.

Previously we covered…

  • Determining the enthalpy and entropy of a system using Hess’s law
  • Using Gibbs Free Energy to calculate the following:
    • Entropy
    • Enthalpy
    • Solving for missing values
    • Determining spontaneity

Le Châtelier’s Principle

Le Châtelier’s principle states that if a change (or stress) is applied to a system at equilibrium, the position of equilibrium will shift in a direction so as to reduce that change. Therefore:

  • If a reactant is added to a reaction at equilibrium, the reaction will produce more products.
  • If a reactant is removed from a reaction at equilibrium, the reaction will shift in the direction which forms more of that reactant.
  • If a product is removed from a reaction at equilibrium, the reaction will shift to form more of that product.
  • If a product is added to a reaction at equilibrium, the reaction will shift to make more of the reactants.
  • An increase of temperature for an endothermic reaction shifts equilibrium towards the products.
  • An increase of temperature for an exothermic reaction shifts equilibrium towards the reactants.
  • When only gases are present, a decrease in volume will cause the equilibrium to shift toward the smaller total number of moles of gases. Remember that to decrease the volume, the pressure must be increased.

As an example, let’s look at the following reaction:

N2(g)+ 3H2(g) ↔ 2NH3(g)+ energy

The changes shown below will result in the corresponding shift.

Change
Shift
Addition of N2(g)
 Towards the products (right)
Removal of N2(g)
 Towards the reactants (left)
Addition of H2 (g)
 Towards the products (right)
Removal of H2 (g)
 Towards the reactants (left)
Addition of NH3(g)
 Towards the reactants (left)
Removal of NH3 (g)
 Towards the products (right)
Addition of energy by heating
 Towards the reactants (left)
Removal of energy by cooling
 Towards the products (right)
Decrease the volume
 Toward the products (right)
Increase the volume
 Toward the reactants (left)

Question

Determine the shift in equilibrium when more carbon dioxide is added to the following reaction.
Energy + CaCO3(s) ↔ CaO(s) + CO2(g)

  1. The reaction will shift toward the products
  2. The reaction will shift toward the reactants

Reveal Answer

The correct answer is B. When more carbon dioxide is added there are more products and the change in equilibrium will shift toward the reactants

Dynamic Equilibrium

When a reaction has the same rate going forward (producing products) as going in reverse (the products reverting to the reactants) the reaction is said to be in equilibrium.

We can calculate the equilibrium constant of a reaction given the concentrations of products and reactants at equilibrium.

First we use the generic equation to set up the equilibrium expression:

aA + bB ↔ cC + dD

The lower case letters represent the coefficients used to balance the equation and the capital letters the substances themselves.

The equilibrium expression is:

The values that are used for the substances are their concentrations measured in molarity.

Sample Problem

Calculate the equilibrium constant under the following conditions:

N2(g) + 3Cl2(g) ↔ 2NCl3(g)

The equilibrium concentrations were found to be NCl3(g) = 0.23 M, N2(g) = 0.12 M, and Cl2(g) = 0.13 M

placeholder.png

Question

At a particular temperature, NO gas reacts with chlorine gas to produce gaseous NOCl. When the reaction reaches equilibrium the concentrations of substances present are NO = 0.56M; Cl2 = 0.34M; and NOCl = 0.012 M. Calculate the value of the equilibrium constant at this temperature.

  1. 1.35 × 10-3 / M
  2. 6.40 × 10-2 / M
  3. 1.59 × 101 / M
  4. 7.40 × 102 / M

Reveal Answer

The correct answer is A.

2NO (g) + Cl2(g) ↔ 2NOCl(g)

Question

At a particular temperature, hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas. Find the value of the equilibrium constant if the following concentrations are found at equilibrium. H2 = 0.0123 M; F2 = 0.0342 M and HF = 0.00291 M.

  1. 2.12 × 10-2
  2. 1.37 × 10-1
  3. 4.71 × 101
  4. 7.27 × 101

Reveal Answer

The correct answer is A.

H2 (g) + F2(g) ↔ 2HF(g)

Another way to find the equilibrium constant of a reaction is to use the equation

e = 2.718, ΔG is free energy, R is the ideal gas law constant 8.314 J/mol·K (or 0.008314 kJ/mol·K), and T is temperature in Kelvin.

Sample Problem

Using the equation above, solve for the equilibrium constant when the value for free energy is 1.5 × 102 kJ at a temperature of 38°C for the reaction:

N2(g) + 3Cl2(g) ↔ 2NCl3(g)

Keq = 6.57 × 10-26

Question

Determine the equilibrium constant for the following reaction when the value for free energy is  2.3×103 kJ and is at a temperature of 1432 K.

2NO (g) + Cl2 (g) ↔ 2NOCl (g)

  1. 1.26 × 10 -84
  2. 1.93 × 10 -1
  3. 8.24 × 10 -1
  4. 1.93 × 10 -58

Reveal Answer

The correct answer is A. Option C is arrived at by neglecting to put R into kJ. Choice D, does not raise e to the specified value and does not convert R into kJ. Option B does not raise the value of e to the power, but is merely the power itself.



Keq=e—193.186
Keq = 1.26 x 10-84

Question

Determine the equilibrium constant for the following reaction that has free energy of -1.34×102 kJ at a temperature of 357 K.

H2(g) + F2(g) ↔ 2HF(g)

  1. 2.47 × 10-20
  2. 1.05 × 101
  3. 4.51 × 102
  4. 4.05 × 1019

Reveal Answer

The correct answer is D. Option C is simply the power to which e is to be raised. Choice A is arrived at by neglecting to make the value of ΔG negative (since the value is negative, it should become positive ). Choice B is obtained by failing to convert the value of R into kJ.



Keq = e45.146
Keq = 4.05 × 1019

⬅ Previous Lesson Workshop Index Next Lesson ➡

Back to Top