Uses of Equilibrium Constants Revision Notes for HSC SSCE Chemistry

Uses of Equilibrium Constants

Understanding what Keq tells us

The equilibrium constant ( $K_{eq}$ ) provides valuable information about the composition of a reaction mixture at equilibrium. By examining the size of $K_{eq}$ , we can determine whether products or reactants predominate in the equilibrium mixture.

Since products appear in the numerator of the equilibrium expression, a large value of $K_{eq}$ indicates high product concentrations compared to reactant concentrations. This allows us to make the following interpretations:

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Interpreting the Magnitude of $K_{eq}$ :

Large equilibrium constant ( $K_{eq} > 10^3$ ):

The equilibrium position lies to the right
Products are strongly favoured
The concentration of products is significantly greater than the concentration of reactants
Most of the reactants have been converted to products

Small equilibrium constant ( $K_{eq} < 10^{-3}$ ):

The equilibrium position lies to the left
Reactants are strongly favoured
The concentration of reactants is significantly greater than the concentration of products
Very little product has formed

Equilibrium constant close to 1 ( $K_{eq}$ between 0.1 and 10):

Neither side is strongly favoured
Both reactants and products are present in significant amounts at equilibrium
The system contains substantial concentrations of both species

Two main applications of equilibrium constants

Equilibrium constants serve two important purposes in chemistry:

Determining whether a system has reached equilibrium - by calculating the reaction quotient ( $Q$ ) and comparing it to $K_{eq}$
Calculating unknown equilibrium concentrations - when we know $K_{eq}$ and have information about other species in the system

Checking if equilibrium has been reached

The reaction quotient (Q)

The reaction quotient ( $Q$ ) has the same mathematical form as the equilibrium constant, but it can be calculated at any point during a reaction, not just at equilibrium. We calculate $Q$ using the concentrations present in the system at a specific moment.

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Comparing Q and $K_{eq}$ to Predict Reaction Direction:

If $Q = K_{eq}$ :

The reaction has reached equilibrium
No net change will occur
The concentrations will remain constant

If $Q < K_{eq}$ :

The reaction has not reached equilibrium
The forward reaction is favoured
The system will shift to the right to produce more products
The numerator (product concentrations) needs to increase

If $Q > K_{eq}$ :

The reaction has not reached equilibrium
The reverse reaction is favoured
The system will shift to the left to produce more reactants
The numerator (product concentrations) needs to decrease

Worked example: The Haber process

The Haber process is industrially important for synthesising ammonia from hydrogen and nitrogen:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

At $427°C$ , the equilibrium constant for this reaction is $0.26$ .

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Worked Example: Determining Equilibrium Status in the Haber Process

Problem: A $4.00 \text{ L}$ flask contains $1.0 \text{ mol}$ of ammonia, $10.0 \text{ mol}$ of hydrogen, and $15.0 \text{ mol}$ of nitrogen. Is the system at equilibrium? If not, which direction will it proceed?

Solution:

Step 1: Convert amounts to concentrations

$[NH_3] = \frac{1.0 \text{ mol}}{4.00 \text{ L}} = 0.25 \text{ mol L}^{-1}$

$[H_2] = \frac{10.0 \text{ mol}}{4.00 \text{ L}} = 2.5 \text{ mol L}^{-1}$

$[N_2] = \frac{15.0 \text{ mol}}{4.00 \text{ L}} = 3.25 \text{ mol L}^{-1}$

Step 2: Write the expression for the reaction quotient

$Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Step 3: Calculate Q using the current concentrations

$Q = \frac{(0.25)^2}{(3.25) \times (2.5)^3} = 1.2 \times 10^{-3}$

Step 4: Compare Q to $K_{eq}$

Since $1.2 \times 10^{-3} \neq 0.26$ , the reaction has not reached equilibrium.

Since $Q < K_{eq}$ , the reaction quotient is too small. To increase $Q$ , the numerator (ammonia concentration) must increase. Therefore, the forward reaction is favoured, and the system will shift to the right to produce more ammonia.

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Exam Tip:

Always compare Q and $K_{eq}$ carefully. Remember that when $Q < K_{eq}$ , the system needs more products, so it shifts right (forward). When $Q > K_{eq}$ , the system needs more reactants, so it shifts left (reverse).

Memory Aid: "Q too low? Forward we go!" and "Q too high? Reverse is nigh!"

Calculating equilibrium concentrations

When we know the equilibrium constant and the equilibrium concentration of at least one species, we can calculate the equilibrium concentrations of other species in the reaction mixture.

Worked example: Phosgene decomposition

Phosgene gas ( $COCl_2$ ) decomposes according to the equation:

$COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)$

The equilibrium constant for this reaction at $1000 \text{ K}$ is $0.40$ . Initially, only phosgene was present. At equilibrium, the concentration of carbon monoxide is $0.24 \text{ mol L}^{-1}$ .

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Worked Example: Calculating Equilibrium Concentrations

Problem: Calculate the equilibrium concentration of phosgene.

Solution:

Step 1: Write the equilibrium expression

$K_{eq} = \frac{[CO][Cl_2]}{[COCl_2]}$

Step 2: Determine the concentration of chlorine

Since only phosgene was initially present and the stoichiometry shows that carbon monoxide and chlorine are produced in a $1:1$ ratio, they must have equal concentrations at equilibrium:

$[Cl_2] = [CO] = 0.24 \text{ mol L}^{-1}$

Step 3: Rearrange the equilibrium expression to solve for phosgene concentration

$[COCl_2] = \frac{[CO][Cl_2]}{K_{eq}}$

Step 4: Substitute the known values and calculate

$[COCl_2] = \frac{(0.24) \times (0.24)}{0.40} = 0.14 \text{ mol L}^{-1}$

The equilibrium concentration of phosgene is $0.14 \text{ mol L}^{-1}$ (to two significant figures).

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Exam Tip:

Pay careful attention to stoichiometric ratios. When only one reactant is initially present, you can often use the balanced equation to relate the concentrations of different products.

Practice problem: Dinitrogen tetroxide decomposition

Dinitrogen tetroxide dissociates into nitrogen dioxide in chloroform solution:

$N_2O_4 \rightleftharpoons 2NO_2$

The concentrations at equilibrium at $298 \text{ K}$ were measured using spectrophotometry for three different mixtures:

Mixture	$[N_2O_4]$ (mol L $^{-1}$ )	$[NO_2]$ (mol L $^{-1}$ )
A	0.129	0.00117
B	0.324	0.00185
C	0.778	0.00284

Task: Calculate the equilibrium constant for each mixture and discuss the reliability of the data.

For each mixture, use the equilibrium expression:

$K_{eq} = \frac{[NO_2]^2}{[N_2O_4]}$

This exercise demonstrates how experimental measurements can verify the constancy of $K_{eq}$ under the same conditions.

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Key Points to Remember:

The magnitude of $K_{eq}$ indicates equilibrium position: Large values ( $>10^3$ ) favour products; small values ( $<10^{-3}$ ) favour reactants; values near $1$ indicate significant amounts of both.
Use the reaction quotient (Q) to check equilibrium status: Calculate $Q$ using current concentrations and compare to $K_{eq}$ . If $Q < K_{eq}$ , the reaction shifts right (forward); if $Q > K_{eq}$ , it shifts left (reverse); if $Q = K_{eq}$ , the system is at equilibrium.
$K_{eq}$ enables concentration calculations: When you know $K_{eq}$ and some equilibrium concentrations, you can calculate unknown equilibrium concentrations using the equilibrium expression.
Stoichiometry is crucial: Always use the balanced chemical equation to relate the concentrations of different species, especially when determining how concentrations of products relate to each other.
Remember the Q-K comparison: A helpful memory aid is "Q too low? Forward we go!" (when $Q < K_{eq}$ , shift right) and "Q too high? Reverse is nigh!" (when $Q > K_{eq}$ , shift left).

Uses of Equilibrium Constants (HSC SSCE Chemistry): Revision Notes