# Equilibrium Constant, K, and Temperature Tutorial

## Key Concepts

• K is the symbol given to the equilibrium constant for a chemical reaction.1

For the general reaction:

reactants ⇋ products     ΔH = ? kJ mol-1

Kc can be represented as :

 Kc = [products][reactants]

• The value of the equilibrium constant, K, for a given reaction is dependent on temperature.
• If ΔH is positive, reaction is endothermic, then:
(a) K increases as temperature increases

(b) K decreases as temperature decreases

• If ΔH is negative, reaction is exothermic, then:
(a) K decreases as temperature increases

(b) K increases as temperature decreases

No ads = no money for us = no free stuff for you!

## Introduction

Imagine a reaction in which the gaseous reactants A(g) and B(g) react in a sealed vessel to produce the gaseous products C(g) at a constant temperature.
The result is an equilibrium reaction which can be represented by the following chemical equation:

reactants products
A(g) + B(g) C(g)

The mass-action expression for this reaction is:

 Q = [C(g)]     [A(g)][B(g)]

At equilibrium the value of the mass-expression is a constant, that is,

Q = a constant

and this cosntant is known as the equilibrium constant, and is given the symbol K, so at equilibrium:

 Q = [C(g)]     [A(g)][B(g)] = a constant = K

At constant temperature, the equilibrium concentrations of A(g), B(g) and C(g) do not appear to change because the rate of the forward reaction is the same as the rate of the reverse reaction.

Do you know this?

Play the game now!

## Effect of Heating on the Value of the Equilibrium Constant, K

What happens if we heat the reaction mixture after it reaches equilibrium?

Remember that chemical reactions themselves either absorb energy (endothermic reactions) or release energy (exothermic reactions).

If our reaction is endothermic, it absorbs energy, so energy (enthalpy, ΔH) can be considered as a reactant in the chemical equilibrium equation:

reactants products
A(g) + B(g) + ΔH C(g)

When the equilibrium mixture is heated, by Le Chatelier's Principle, the equilibrium position will shift to the right to consume some of the additional heat.

This means that when the reaction mixture reaches equilibrium again, the concentration of each reactant will have decreased by an amount x, while the concentration of the product will have increased by an amount x.

We can summarise these changes in concentration of each species in a R.I.C.E. (or I.C.E.) table as shown below:

reactants ⇋ products Reaction: Initial concentration:(before heating) Change in concentration:(heated to new temperature) A(g) + B(g) + ΔH ⇋ C(g) [A(g)] [B(g)] [C(g)] -x -x +x [A(g)] - x [B(g)] - x [C(g)] + x

Let's compare the relative values of the equilibrium constant before and after heating the reaction mixture:

Before Heating After Heating
Equilibrium constants:
 K = [C(g)]     [A(g)][B(g)]
 K' = [C(g)] + x     ([A(g)] - x)([B(g)] - x)

We can see that K' will be a larger number than K because:

• the value of the numerator has increased (from [C(g)] to [C(g)] + x)
• and the value of the denominator has decreased (from [A(g)][B(g)] to {([A(g)] - x)([B(g)] - x)}
• and a larger number divided by a smaller number results in a larger number

If a reaction is endothermic, the value of the equilibrium constant increases when the reaction mixture is heated.

Endothermic reaction: increasing temperature, increases the value of K

Compare what happens to the value of the equilibrium constant when an exothermic reaction is heated.

An exothermic reaction releases energy, so energy (enthalpy, ΔH) can be considered as a product in the chemical equilibrium equation:

reactants products
A(g) + B(g) C(g) + ΔH

When the equilibrium mixture is heated, by Le Chatelier's Principle, the equilibrium position will shift to the left to consume some of the additional heat.

This means that when the reaction mixture reaches equilibrium again, the concentration of each reactant will have increased by an amount x, while the concentration of the product will have decreased by an amount x.
We can represent these changes in a R.I.C.E. Table as shown below:

reactants ⇋ products Reaction: Initial concentration:(before heating) Change in concentration:(heated to new temperature) A(g) + B(g) ⇋ C(g) + ΔH [A(g)] [B(g)] [C(g)] +x +x -x [A(g)] + x [B(g)] + x [C(g)] - x

Let's compare the relative values of the equilibrium constant before and after heating the reaction mixture:

Before Heating After Heating
Equilibrium constants:
 K = [C(g)]     [A(g)][B(g)]
 K' = [C(g)] - x     ([A(g)] + x)([B(g)] + x)

We can see that K' will be a smaller number than K because:

• the value of the numerator has decreased from [C(g)] to ([C(g)] - x)
• and the value of the denominator has increased from [A(g)][B(g)] to {([A(g)] + x)([B(g)] + x)}
• and a smaller number divided by a larger number results in a smaller number

If a reaction is exothermic, the value of the equilibrium constant decreases when the reaction mixture is heated.

Exothermic reaction: increasing temperature, decreases the value of K

Do you understand this?

Take the test now!

## Effect of Cooling on the Value of the Equilibrium Constant, K

If our reaction is endothermic, it absorbs energy, so energy (enthalpy, ΔH) can be considered as a reactant in the chemical equilibrium equation:

reactants products
A(g) + B(g) + ΔH C(g)

When the equilibrium mixture is cooled, that is heat is removed, by Le Chatelier's Principle, the equilibrium position will shift to the left to produce some additional heat.

This means that when the reaction mixture reaches equilibrium again, the concentration of each reactant will have increased by an amount x, while the concentration of the product will have decreased by an amount x.

We can summarise these changes in concentration of each species in a R.I.C.E. (or I.C.E.) table as shown below:

reactants ⇋ products Reaction: Initial concentration:(before cooling) Change in concentration:(cooled to new temperature) A(g) + B(g) + ΔH ⇋ C(g) [A(g)] [B(g)] [C(g)] +x +x -x [A(g)] + x [B(g)] + x [C(g)] - x

Let's compare the relative values of the equilibrium constant before and after cooling the reaction mixture:

Before Cooling After Cooling
Equilibrium constants:
 K = [C(g)]     [A(g)][B(g)]
 K' = [C(g)] - x     ([A(g)] + x)([B(g)] + x)

We can see that K' will be a smaller number than K because:

• the value of the numerator has decreased from [C(g)] to ([C(g)] - x)
• and the value of the denominator has increased from [A(g)][B(g)] to {([A(g)] + x)([B(g)] + x)}
• and a smaller number divided by a larger number results in a smaller number

If a reaction is endothermic, the value of the equilibrium constant decreases when the reaction mixture is cooled.

Endothermic reaction: decreasing temperature, decreases the value of K

If our reaction is exothermic, it releases energy, so energy (enthalpy, ΔH) can be considered as a product in the chemical equilibrium equation:

reactants products
A(g) + B(g) C(g) + ΔH

When the equilibrium mixture is cooled, that is heat is removed, by Le Chatelier's Principle, the equilibrium position will shift to the right to produce some additional heat.

This means that when the reaction mixture reaches equilibrium again, the concentration of each reactant will have decreased by an amount x, while the concentration of the product will have increased by an amount x.

We can summarise these changes in concentration of each species in a R.I.C.E. (or I.C.E.) table as shown below:

reactants ⇋ products Reaction: Initial concentration:(before cooling) Change in concentration:(cooled to new temperature) A(g) + B(g) ⇋ C(g) + ΔH [A(g)] [B(g)] [C(g)] -x -x +x [A(g)] - x [B(g)] - x [C(g)] + x

Let's compare the relative values of the equilibrium constant before and after cooling the reaction mixture:

Before Cooling After Cooling
Equilibrium constants:
 K = [C(g)]     [A(g)][B(g)]
 K' = [C(g)] + x     ([A(g)] - x)([B(g)] - x)

We can see that K' will be a larger number than K because:

• the value of the numerator has increased (from [C(g)] to [C(g)] + x)
• and the value of the denominator has decreased (from [A(g)][B(g)] to {[A(g)] - x}{[B(g)] - x})
• and a larger number divided by a smaller number results in a larger number

If a reaction is exothermic, the value of the equilibrium constant increases when the reaction mixture is cooled.

Exothermic reaction: decreasing temperature, increases the value of K

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

## Worked Example: Predict the Value for an Equilibrium Constant, K, at a Different Temperature

Question: The decomposition of N2O4(g) to produce NO2(g) is an endothermic chemical reaction which can be represented by the following chemical equation:

N2O4(g) ⇋ 2NO2(g)

At 25°C the value of the equilibrium constant, Kc is 4.7 × 10-3.

Predict whether the equilibrium constant for this reaction at 100°C will be greater than, less than, or equal to 4.7 × 10-3.

Solution: (based on the StoPGoPS approach to problem solving)

STOP STOP! State the Question.
What is the question asking you to do?

For 100oC decide whether

Kc > 4.7 × 10-3

or Kc < 4.7 × 10-3

or Kc = 4.7 × 10-3

PAUSE PAUSE to Prepare a Game Plan

1. What data have you been given?

Reaction is said to be endothermic so:

N2O4(g) + ΔH ⇋ 2NO2(g)

At 25°C, Kc = 4.7 × 10-3

2. What is the relationship between what you have been given and what you need to find?

Endothermic reaction: K increases as temperature increases.

GO GO with the Game Plan
Determine the relative value for Kc at 100oC

Temperature has increased: 100°C is a higher temperature than 25°C

Therefore, Kc for this endothermic reaction will increase.

Kc at 100°C will be greater than Kc at 25°C

At 100°C, Kc > 4.7 × 10-3

PAUSE PAUSE to Ponder Plausibility

reactants products Kc
Reaction: N2O4(g) + ΔH 2NO2(g)
Initial concentration:
(before heating)
[N2O4(g)]     [NO2(g)]
 K = [NO2]2[N2O4]
Change in concentration:
(heated to new temperature)
-x     +2x
Equilibrium concentration:
(at new higher temperature)
[N2O4(g)] - x     [NO2(g)] + 2x
 K' = ([NO2]+ 2x) 2([N2O4]- x)

K' will be larger than K because:

• the value of the numerator has increased from [NO2]2 to ([NO2]+ 2x) 2
• and the value of the denominator has decreased from [N2O4] to ([N2O4]- x)
• and a larger number divided by a smaller number will give a larger number

K' will be greater than 4.7 × 10-3 so our answer is plausible.

STOP STOP! State the Solution
State your solution to the problem.

The value of the equilibrium constant will be greater than 4.7 × 10-3 at 100°C.

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

## Worked Example: Use Equilibrium Constant Values at Different Temperatures to Predict Relative Value of ΔH

Question: Consider the synthesis of I3-(aq) from the reactants I2(aq) and I-(aq) as shown by the following chemical equation:

I2(aq) + I-(aq) ⇋ I3-(aq)

The reaction was carried out at a number of different temperatures and the equilibrium constant, Kc, was calculated for each temperature as shown in the table below:

Temperature
/ °C
Kc
0 1360
25 723
50 370
100 190

Is this reaction exothermic or endothermic?

Solution: (based on the StoPGoPS approach to problem solving)

STOP STOP! State the Question.
What is the question asking you to do?

State whether the reaction is either
(a) exothermic
or
(b) endothermic

PAUSE PAUSE to Prepare a Game Plan

1. What data have you been given?

Reaction: I2(aq) + I-(aq) ⇋ I3-(aq)

Variation of Kc with Temperature
Temperature
/ oC
Kc
0 1360
25 723
50 370
100 190

2. What is the relationship between what you have been given and what you need to find?

Endothermic reaction: Kc increases as temperature increases

Exothermic reaction: Kc decreases as temperature increases

GO GO with the Game Plan
Determine whether the reaction is exothermic or endothermic

From the data in the table, as temperature increases from 0°C to 100°C, the value of Kc decreases from 1360 to 190.

As temperature increases, Kc decreases.

This reaction is exothermic.

PAUSE PAUSE to Ponder Plausibility

Assume the reaction is exothermic and predict the effect of increasing temperature on the value of Kc

reactants products Kc
Reaction: I2(aq) + I-(aq) I3-(aq) + ΔH
Initial concentration:
(before heating)
[I2(aq)]   [I-(aq)]   [I3-(aq)]
 K = [I3-(aq)]     [I2(aq)][I-(aq)]
Change in concentration:
(heated to new temperature)
+x   +x   -x
Equilibrium concentration:
(at new higher temperature)
[I2(aq)] + x   [I-(aq)] + x   [I3-(aq)] - x
 K' = ([I3-(aq)] - x)         {([I2(aq)] + x)([I-(aq)] + x)}

We can see that K' will be a smaller number than K because:

• the value of the numerator has decreased from [I3-(aq)] to [I3-(aq)] - x
• and the value of the denominator has increased from [I2(aq)][I-(aq)] to ([I2(aq)] + x)([I-(aq)] + x)
• and a smaller number divided by a larger number results in a smaller number

For an exothermic reaction the value of the equilibrium constant decreases when temperature increases, which is what we saw in the data in the question, so we are confident are answer is correct.

STOP STOP! State the Solution
State your solution to the problem.

I2(aq) + I-(aq) ⇋ I3-(aq) is an exothermic reaction.

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

1. Although this discussion will concern Kc, the same logic can be applied to KP, for this reason we have used K rather than Kc for generalisations regarding the extent of reaction and the magnitude of the equilibrium constant.