Equilibrium Expressions and Concentrations: HSC Chemistry Practice Exam

This original tutorgum practice paper tests equilibrium expressions and concentration calculations for the NSW HSC Chemistry 2025 syllabus scope supplied for this assessment. It covers constructing equilibrium expressions, calculating equilibrium constants, predicting reaction direction, reciprocal equilibrium constants, and solving equilibrium concentrations using stoichiometric changes.

The paper is designed to test both your calculation skills and your understanding of what the numbers mean. It includes a quadratic equilibrium calculation where you must identify and reject a physically impossible solution.

This is an original tutorgum practice paper. It is not an official, endorsed, past, or predicted NESA examination paper.

01Exam instructions

Suggested time: 5 minutes reading time and 65 minutes working time (70 minutes total).

Total marks: 40

Permitted equipment:

• Scientific calculator.
• Pens, pencils, and ruler.
• No notes or external resources.

General instructions:

1. Attempt every question.
2. Show all working for calculations. A correct numerical answer without supporting working may not receive full marks.
3. Give concentrations in mol L\(^{-1}\).
4. Round final answers sensibly to the precision of the supplied data. Avoid rounding intermediate calculations.
5. Use \(K_c\) for the equilibrium constant and \(Q_c\) for the reaction quotient, calculated using concentrations at a particular instant.
6. Unless stated otherwise, assume temperature and volume remain constant.
7. All reactions are homogeneous reactions in aqueous solution. Letter symbols represent distinct dissolved species, and the equations are stipulated to be balanced.
8. In an ICE table, \(x\) represents a concentration change in mol L\(^{-1}\). Apply the stoichiometric coefficients when calculating each species’ change.

Assessed content

02Questions

Question 1 (6 marks)

Consider the following homogeneous equilibrium reaction:

\[ \ce{2A(aq) + B(aq) <=> C(aq) + 2D(aq)} \]

At a particular temperature, the equilibrium concentrations are:

(a) Write the equilibrium constant expression, \(K_c\), for this reaction. (2 marks)

(b) Calculate the equilibrium constant for the reaction using the supplied concentrations. (4 marks)

Question 2 (5 marks)

Consider the reaction:

\[ \ce{X(aq) + 2Y(aq) <=> XY2(aq)} \]

At a particular temperature, the equilibrium constant is \(K_c=0.80\).

A mixture at the same temperature has the following concentrations at a particular instant:

(a) Write the expression for the reaction quotient, \(Q_c\). (1 mark)

(b) Calculate the reaction quotient for the mixture. (2 marks)

(c) Predict the net direction in which the reaction will proceed to reach equilibrium. Justify your answer using the relationship between \(Q_c\) and \(K_c\). (2 marks)

Question 3 (10 marks)

Consider the homogeneous equilibrium:

\[ \ce{A(aq) + B(aq) <=> C(aq)} \]

A reaction mixture initially contains:

\[ [A]_0=1.00\text{ mol L}^{-1} \]

\[ [B]_0=0.50\text{ mol L}^{-1} \]

\[ [C]_0=0.10\text{ mol L}^{-1} \]

At the temperature of the reaction, the equilibrium constant is:

\[ K_c=0.40 \]

(a) Calculate the initial reaction quotient, \(Q_c\), and determine the initial net direction of the reaction. (2 marks)

(b) Construct an ICE table for the reaction. Define \(x\) as the concentration change in the forward direction and write expressions for all three equilibrium concentrations. (3 marks)

(c) Substitute your equilibrium concentrations into the equilibrium constant expression. Form a quadratic equation, solve for \(x\), and identify the physically valid solution. Explain why the other solution must be rejected. (3 marks)

(d) Calculate the equilibrium concentration of each species. (2 marks)

Question 4 (6 marks)

Consider the homogeneous equilibrium:

\[ \ce{R(aq) + 2S(aq) <=> T(aq)} \]

The equilibrium constant for the forward reaction is:

\[ K_{c,\text{forward}}=2.5 \]

(a) Write the equilibrium constant expression for the forward reaction. (1 mark)

(b) Write the balanced chemical equation for the reverse reaction. (1 mark)

(c) Write the equilibrium constant expression for the reverse reaction. (1 mark)

(d) Calculate the equilibrium constant for the reverse reaction. (1 mark)

(e) At equilibrium, the concentrations of R and S are:

\[ [R]=0.20\text{ mol L}^{-1} \]

\[ [S]=0.40\text{ mol L}^{-1} \]

Calculate the equilibrium concentration of T. (2 marks)

Question 5 (8 marks)

Consider the homogeneous equilibrium:

\[ \ce{2G(aq) <=> H(aq) + J(aq)} \]

A reaction mixture initially contains:

\[ [G]_0=0.60\text{ mol L}^{-1} \]

\[ [H]_0=0.10\text{ mol L}^{-1} \]

\[ [J]_0=0.10\text{ mol L}^{-1} \]

The reaction proceeds in the forward direction before reaching equilibrium. At this temperature:

\[ K_c=0.25 \]

(a) Construct an ICE table for the reaction. Define \(x\) as the concentration change in the forward direction and express each equilibrium concentration in terms of \(x\). (2 marks)

(b) Substitute your expressions into the equilibrium constant equation and solve for \(x\). (3 marks)

(c) Calculate the equilibrium concentrations of G, H, and J. (3 marks)

Question 6 (5 marks)

Consider the homogeneous equilibrium:

\[ \ce{E(aq) + 2F(aq) <=> EF2(aq)} \]

At a particular temperature:

\[ K_c=2.0 \]

(a) Write the equilibrium constant expression for this reaction. (1 mark)

(b) At equilibrium, the concentrations of F and EF\(_2\) are:

\[ [F]=0.25\text{ mol L}^{-1} \]

\[ [EF_2]=0.10\text{ mol L}^{-1} \]

Calculate the equilibrium concentration of E. (2 marks)

(c) A different mixture at the same temperature has the following concentrations:

\[ [E]=0.40\text{ mol L}^{-1} \]

\[ [F]=0.25\text{ mol L}^{-1} \]

\[ [EF_2]=0.10\text{ mol L}^{-1} \]

Calculate \(Q_c\) for this mixture and predict the net direction in which the reaction will proceed to reach equilibrium. (2 marks)

03Worked solutions and marking guide

Solution 1

(a) Write the equilibrium constant expression. (2 marks)

The general rule for a homogeneous equilibrium expression is:

• Products go in the numerator.
• Reactants go in the denominator.
• Each concentration is raised to the power of its stoichiometric coefficient.

For:

\[ \ce{2A + B <=> C + 2D} \]

The expression is:

\[ \boxed{K_c=\frac{[C][D]^2}{[A]^2[B]}} \]

The coefficient of A is 2, so \([A]\) is squared. The coefficient of D is also 2, so \([D]\) is squared.

The coefficients of B and C are both 1, so neither concentration needs an explicit power.

Marking guide:

• Correct products in the numerator and reactants in the denominator: 1 mark.
• Correct stoichiometric powers: 1 mark.

(b) Calculate the equilibrium constant. (4 marks)

Substitute the supplied equilibrium concentrations:

\[ K_c=\frac{[C][D]^2}{[A]^2[B]} \]

\[ K_c=\frac{0.15(0.40)^2}{(0.30)^2(0.20)} \]

Evaluate the powers:

\[ K_c=\frac{0.15(0.16)}{0.09(0.20)} \]

\[ K_c=\frac{0.024}{0.018} \]

Therefore:

\[ \boxed{K_c=1.33} \]

Marking guide:

• Correct substitution of equilibrium concentrations: 1 mark.
• Correct evaluation of squared concentrations: 1 mark.
• Correct numerator and denominator calculation: 1 mark.
• Correct final answer: 1 mark.

Solution 2

(a) Write the reaction quotient expression. (1 mark)

The reaction quotient, \(Q_c\), has the same mathematical form as the equilibrium constant expression.

The difference is that \(Q_c\) uses the concentrations at the instant being examined, rather than necessarily using equilibrium concentrations.

For:

\[ \ce{X + 2Y <=> XY2} \]

The expression is:

\[ \boxed{Q_c=\frac{[XY_2]}{[X][Y]^2}} \]

Marking guide:

• Correct expression with the appropriate species and powers: 1 mark.

(b) Calculate the reaction quotient. (2 marks)

Substitute the current concentrations:

\[ Q_c=\frac{0.050}{(0.50)(0.20)^2} \]

Evaluate the square:

\[ Q_c=\frac{0.050}{(0.50)(0.040)} \]

\[ Q_c=\frac{0.050}{0.020} \]

Therefore:

\[ \boxed{Q_c=2.5} \]

Marking guide:

• Correct substitution and evaluation of the squared concentration: 1 mark.
• Correct numerical answer: 1 mark.

(c) Predict the net reaction direction. (2 marks)

Compare the calculated reaction quotient with the equilibrium constant:

\[ Q_c=2.5 \]

\[ K_c=0.80 \]

Therefore:

\[ Q_c>K_c \]

When \(Q_c>K_c\), the mixture contains too much product relative to the equilibrium composition.

The reaction will proceed in the reverse direction, consuming XY\(_2\) and producing X and Y.

\[ \boxed{\text{Net reaction proceeds to the left.}} \]

Marking guide:

• Correct comparison showing \(Q_c>K_c\): 1 mark.
• Correct prediction and explanation that the reverse reaction consumes product: 1 mark.

Solution 3

(a) Calculate the initial reaction quotient and predict the direction. (2 marks)

The reaction is:

\[ \ce{A + B <=> C} \]

The reaction quotient expression is:

\[ Q_c=\frac{[C]}{[A][B]} \]

Substitute the initial concentrations:

\[ Q_c=\frac{0.10}{(1.00)(0.50)} \]

\[ \boxed{Q_c=0.20} \]

Compare this with the equilibrium constant:

\[ Q_c=0.20 \]

\[ K_c=0.40 \]

Since:

\[ Q_c<K_c \]

the reaction proceeds in the forward direction to produce more C.

\[ \boxed{\text{Net reaction proceeds to the right.}} \]

Marking guide:

• Correct expression and initial quotient calculation: 1 mark.
• Correct comparison with \(K_c\) and predicted direction: 1 mark.

(b) Construct an ICE table. (3 marks)

An ICE table tracks the initial concentrations, the changes during the reaction, and the equilibrium concentrations.

ICE stands for:

• Initial
• Change
• Equilibrium

Because the reaction proceeds in the forward direction:

\[ \ce{A + B -> C} \]

A and B are consumed, while C is produced.

The stoichiometric ratio is 1:1:1, so the concentration changes are \(-x\), \(-x\), and \(+x\).

Therefore:

\[ [A]_{\text{eq}}=1.00-x \]

\[ [B]_{\text{eq}}=0.50-x \]

\[ [C]_{\text{eq}}=0.10+x \]

Marking guide:

• Correct stoichiometric concentration changes: 1 mark.
• Correct equilibrium expressions for A and B: 1 mark.
• Correct equilibrium expression for C: 1 mark.

(c) Form and solve the quadratic equation. (3 marks)

The equilibrium constant expression is:

\[ K_c=\frac{[C]}{[A][B]} \]

Substitute the equilibrium concentrations from the ICE table:

\[ 0.40=\frac{0.10+x}{(1.00-x)(0.50-x)} \]

First, expand the denominator:

\[ (1.00-x)(0.50-x) \]

\[ =0.50-1.50x+x^2 \]

Substitute this into the equation:

\[ 0.40(0.50-1.50x+x^2)=0.10+x \]

Expand:

\[ 0.20-0.60x+0.40x^2=0.10+x \]

Rearrange into quadratic form:

\[ 0.40x^2-1.60x+0.10=0 \]

Multiply by 10:

\[ 4x^2-16x+1=0 \]

Apply the quadratic formula:

\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]

Here:

\[ a=4,\quad b=-16,\quad c=1 \]

Therefore:

\[ x=\frac{16\pm\sqrt{256-16}}{8} \]

\[ x=\frac{16\pm\sqrt{240}}{8} \]

The two solutions are:

\[ x=0.0635 \]

or

\[ x=3.94 \]

We must now check which solution is physically possible.

The initial concentration of B is only \(0.50\text{ mol L}^{-1}\). Since B is consumed by \(x\), its equilibrium concentration must be:

\[ [B]_{\text{eq}}=0.50-x \]

If \(x=3.94\):

\[ [B]_{\text{eq}}=0.50-3.94 \]

\[ [B]_{\text{eq}}=-3.44\text{ mol L}^{-1} \]

A negative concentration is physically impossible.

Therefore, the only valid solution is:

\[ \boxed{x=0.0635\text{ mol L}^{-1}} \]

Marking guide:

• Correct substitution into the equilibrium expression and formation of the quadratic: 1 mark.
• Correct quadratic solutions: 1 mark.
• Correct identification of the physically valid solution, with explanation for rejecting the other root: 1 mark.

(d) Calculate the equilibrium concentrations. (2 marks)

Use the physically valid value:

\[ x=0.0635083 \]

For A:

\[ [A]_{\text{eq}}=1.00-x \]

\[ [A]_{\text{eq}}=1.00-0.0635083 \]

\[ \boxed{[A]_{\text{eq}}=0.936\text{ mol L}^{-1}} \]

For B:

\[ [B]_{\text{eq}}=0.50-x \]

\[ [B]_{\text{eq}}=0.50-0.0635083 \]

\[ \boxed{[B]_{\text{eq}}=0.436\text{ mol L}^{-1}} \]

For C:

\[ [C]_{\text{eq}}=0.10+x \]

\[ [C]_{\text{eq}}=0.10+0.0635083 \]

\[ \boxed{[C]_{\text{eq}}=0.164\text{ mol L}^{-1}} \]

Marking guide:

• Correct equilibrium concentrations for A and B: 1 mark.
• Correct equilibrium concentration for C, with appropriate units and precision: 1 mark.

Solution 4

(a) Write the forward equilibrium expression. (1 mark)

The forward reaction is:

\[ \ce{R + 2S <=> T} \]

Products go in the numerator, and reactants go in the denominator.

The coefficient of S is 2.

Therefore:

\[ \boxed{K_{c,\text{forward}}=\frac{[T]}{[R][S]^2}} \]

Marking guide:

• Correct expression and stoichiometric power: 1 mark.

(b) Write the reverse reaction. (1 mark)

The reverse reaction swaps the reactants and products:

\[ \boxed{\ce{T(aq) <=> R(aq) + 2S(aq)}} \]

Marking guide:

• Correct reverse equation, including stoichiometric coefficients: 1 mark.

(c) Write the reverse equilibrium expression. (1 mark)

For the reverse reaction, T is the reactant, while R and S are products.

Therefore:

\[ \boxed{K_{c,\text{reverse}}=\frac{[R][S]^2}{[T]}} \]

Marking guide:

• Correct reverse equilibrium expression: 1 mark.

(d) Calculate the reverse equilibrium constant. (1 mark)

When a reaction is reversed, its equilibrium constant becomes the reciprocal of the original equilibrium constant.

The relationship is:

\[ K_{c,\text{reverse}}=\frac{1}{K_{c,\text{forward}}} \]

Substitute the given value:

\[ K_{c,\text{reverse}}=\frac{1}{2.5} \]

Therefore:

\[ \boxed{K_{c,\text{reverse}}=0.40} \]

Marking guide:

• Correct reciprocal relationship and numerical answer: 1 mark.

(e) Calculate the equilibrium concentration of T. (2 marks)

Start with the forward equilibrium expression:

\[ K_c=\frac{[T]}{[R][S]^2} \]

Rearrange to make \([T]\) the subject:

\[ [T]=K_c[R][S]^2 \]

Substitute the supplied values:

\[ [T]=2.5(0.20)(0.40)^2 \]

\[ [T]=2.5(0.20)(0.16) \]

\[ [T]=0.080 \]

Therefore:

\[ \boxed{[T]=0.080\text{ mol L}^{-1}} \]

Marking guide:

• Correct rearrangement of the equilibrium expression: 1 mark.
• Correct substitution, calculation, and final concentration with units: 1 mark.

Solution 5

(a) Construct an ICE table. (2 marks)

The reaction is:

\[ \ce{2G <=> H + J} \]

The question states that the reaction proceeds in the forward direction.

For every two moles of G consumed, one mole of H and one mole of J are produced.

Therefore, the concentration changes are:

• G: \(-2x\)
• H: \(+x\)
• J: \(+x\)

The ICE table is:

Therefore:

\[ [G]_{\text{eq}}=0.60-2x \]

\[ [H]_{\text{eq}}=0.10+x \]

\[ [J]_{\text{eq}}=0.10+x \]

Marking guide:

• Correct stoichiometric changes for all three species: 1 mark.
• Correct equilibrium concentration expressions: 1 mark.

(b) Substitute into the equilibrium expression and solve for \(x\). (3 marks)

The equilibrium constant expression is:

\[ K_c=\frac{[H][J]}{[G]^2} \]

Substitute the equilibrium concentrations:

\[ 0.25=\frac{(0.10+x)(0.10+x)} {(0.60-2x)^2} \]

Simplify:

\[ 0.25=\frac{(0.10+x)^2} {(0.60-2x)^2} \]

Take the positive square root of both sides. The concentrations must remain positive, so the ratio is positive:

\[ 0.50=\frac{0.10+x}{0.60-2x} \]

Cross-multiply:

\[ 0.50(0.60-2x)=0.10+x \]

Expand:

\[ 0.30-x=0.10+x \]

Rearrange:

\[ 0.20=2x \]

Therefore:

\[ \boxed{x=0.10\text{ mol L}^{-1}} \]

Marking guide:

• Correct substitution into the equilibrium expression: 1 mark.
• Correct use of the positive square root and rearrangement: 1 mark.
• Correct value of \(x\): 1 mark.

(c) Calculate the equilibrium concentrations. (3 marks)

Use:

\[ x=0.10 \]

For G:

\[ [G]_{\text{eq}}=0.60-2(0.10) \]

\[ \boxed{[G]_{\text{eq}}=0.40\text{ mol L}^{-1}} \]

For H:

\[ [H]_{\text{eq}}=0.10+0.10 \]

\[ \boxed{[H]_{\text{eq}}=0.20\text{ mol L}^{-1}} \]

For J:

\[ [J]_{\text{eq}}=0.10+0.10 \]

\[ \boxed{[J]_{\text{eq}}=0.20\text{ mol L}^{-1}} \]

Check that these values satisfy the original equilibrium expression:

\[ K_c=\frac{(0.20)(0.20)}{(0.40)^2} \]

\[ K_c=\frac{0.040}{0.160} \]

\[ K_c=0.25 \]

The calculated concentrations are consistent with the supplied equilibrium constant.

Marking guide:

• Correct equilibrium concentration of G: 1 mark.
• Correct equilibrium concentrations of H and J: 1 mark.
• Correct units and verification against the given equilibrium constant: 1 mark.

Solution 6

(a) Write the equilibrium constant expression. (1 mark)

The reaction is:

\[ \ce{E + 2F <=> EF2} \]

The product goes in the numerator, and the reactants go in the denominator.

The coefficient of F is 2, so its concentration is squared.

Therefore:

\[ \boxed{K_c=\frac{[EF_2]}{[E][F]^2}} \]

Marking guide:

• Correct equilibrium expression with the correct stoichiometric power: 1 mark.

(b) Calculate the equilibrium concentration of E. (2 marks)

Start with:

\[ K_c=\frac{[EF_2]}{[E][F]^2} \]

We need to find \([E]\), so rearrange:

\[ [E]=\frac{[EF_2]}{K_c[F]^2} \]

Substitute the given values:

\[ [E]=\frac{0.10}{2.0(0.25)^2} \]

Evaluate the square:

\[ [E]=\frac{0.10}{2.0(0.0625)} \]

\[ [E]=\frac{0.10}{0.125} \]

Therefore:

\[ \boxed{[E]=0.80\text{ mol L}^{-1}} \]

Marking guide:

• Correct rearrangement and substitution: 1 mark.
• Correct numerical answer with units: 1 mark.

(c) Calculate the reaction quotient and predict the reaction direction. (2 marks)

The reaction quotient has the same expression as the equilibrium constant:

\[ Q_c=\frac{[EF_2]}{[E][F]^2} \]

Substitute the concentrations of the different mixture:

\[ Q_c=\frac{0.10}{(0.40)(0.25)^2} \]

\[ Q_c=\frac{0.10}{(0.40)(0.0625)} \]

\[ Q_c=\frac{0.10}{0.025} \]

Therefore:

\[ \boxed{Q_c=4.0} \]

Compare this with the equilibrium constant:

\[ Q_c=4.0 \]

\[ K_c=2.0 \]

Since:

\[ Q_c>K_c \]

the mixture contains too much product relative to the equilibrium composition.

The net reaction therefore proceeds in the reverse direction, consuming EF\(_2\) and producing E and F.

\[ \boxed{\text{Net reaction proceeds to the left.}} \]

Marking guide:

• Correct reaction quotient calculation: 1 mark.
• Correct comparison with \(K_c\), including the predicted reverse direction: 1 mark.

End of paper. Total: 40 marks.