How many QCAA Chemistry questions cover Chemical equilibrium systems?

AusGrader has 399 QCAA Chemistry questions on Chemical equilibrium systems, all with instant AI grading and detailed marking feedback.

QCAA Chemistry — Chemical equilibrium systems Questions & Answers

Q1

2020

QCAA

Paper 1

1 mark

Q1

1 mark

A partly filled water bottle is sealed and left on a bench in a room with a constant temperature. After several minutes, it is noted that the water level in the bottle remains constant. In the water bottle, the rate of evaporation is

A

less than the rate of condensation.

B

greater than the rate of condensation.

C

equal to the rate of condensation and equal to zero.

D

equal to the rate of condensation but not equal to zero.

Reveal Answer

A

less than the rate of condensation.

If the rate of evaporation were less than the rate of condensation, the amount of liquid water would increase, causing the water level to rise.

B

greater than the rate of condensation.

If the rate of evaporation were greater than the rate of condensation, the amount of liquid water would decrease, causing the water level to drop.

C

equal to the rate of condensation and equal to zero.

While the rates are equal, they are not zero because molecules are constantly escaping and returning to the liquid surface in a state of dynamic equilibrium.

D

equal to the rate of condensation but not equal to zero.

Correct Answer

A constant water level in a closed system indicates dynamic equilibrium, where the rate of evaporation equals the rate of condensation, and both processes continue to occur simultaneously.

Q38

2021

SCSA

12 marks

Q38

12 marks

Sulfuric acid is manufactured by the Contact process, the steps of which are outlined below.

Step One: Molten sulfur is burned in air at approximately 1000 °C:

S(l) + O_2(g) \rightarrow SO_2(g) + 297 kJ

Step Two: The resulting sulfur dioxide is converted to sulfur trioxide as shown in the following equilibrium reaction. It is conducted at a temperature of about 450 °C with a $V_2O_5$ catalyst at a pressure of between 100 and 200 kPa:

2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g) + 198 kJ

Step Three: The resulting sulfur trioxide is absorbed into sulfuric acid, producing oleum ( $H_2S_2O_7$ ). Water is added to the oleum, producing 18 mol L $^{-1}$ sulfuric acid:

SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l)

H_2S_2O_7(l) + H_2O(l) \rightarrow 2 H_2SO_4(aq)

Use your understanding of collision theory and chemical equilibrium to discuss the reaction conditions for Steps 1 and 2 of the Contact process, given that the aim is to produce the greatest yield in the shortest time. In your discussion, also address economic concerns where appropriate.

Reveal Answer

High temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently. Also, more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases. The vanadium catalyst increases the rate of the forward reaction, and also the rate of the reverse reaction to an equal extent, as it provides an alternative pathway with a lower activation energy. Therefore, a greater proportion of the particles will have sufficient energy to react when they collide. High pressure or concentration has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases. As Step 1 is a combustion reaction, it essentially goes to completion at the high temperature and does not require a catalyst or high pressure. For Step 2, high temperature, high pressure and a catalyst would favour a high rate.

For equilibrium, which is only considered for Step 2, high temperature favours the reverse reaction because it is endothermic, and this decreases the SO $_3$ (g) yield, which is not desired. A low temperature decreases the rate of reaction, which is also not desired. A high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO $_3$ (g) yield which is desired.

Economically, high pressures are costly and dangerous.

Therefore, for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield. A compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.

Marking Criteria

Rates

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.	6

Descriptor

Marks

1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.

6

Equilibrium

Descriptor	Marks
1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.	3

Descriptor

Marks

1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.

3

Economics

Descriptor	Marks
States that high pressures are costly (and dangerous).	1

Compromise

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 2 marks: Identifies that for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield; Identifies that a compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.	2

Q5

2023

QCAA

Paper 1

1 mark

Q5

1 mark

The question refers to the decomposition of hydrogen iodide gas (HI) to produce hydrogen gas ( $\text{H}_2$ ) and iodine gas ( $\text{I}_2$ ) in a sealed 1-litre container.

$2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \quad \Delta H = +53.6 \text{ kJ mol}^{-1}$
Colourless Colourless Purple

Determine the equilibrium expression ( $K_c$ ) for the reaction.

A

$K_c = \frac{[\text{H}_2][\text{I}_2]}{2[\text{HI}]}$

B

$K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}$

C

$K_c = \frac{2[\text{H}]2[\text{I}]}{2[\text{HI}]}$

D

$K_c = \frac{2[\text{H}]2[\text{I}]}{[\text{HI}]^2}$

Reveal Answer

A

$K_c = \frac{[\text{H}_2][\text{I}_2]}{2[\text{HI}]}$

This is incorrect because the stoichiometric coefficient of the reactant (2) must be used as an exponent, not a multiplier. The correct term is $[\text{HI}]^2$ , not $2[\text{HI}]$ .

B

$K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}$

Correct Answer

This is correct based on the law of mass action for the reaction $2\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2$ . The concentration of products is in the numerator, the reactant is in the denominator, and the coefficient 2 for HI becomes an exponent.

C

$K_c = \frac{2[\text{H}]2[\text{I}]}{2[\text{HI}]}$

This is incorrect because it uses atomic species (H, I) instead of the molecular species ( $\text{H}_2$ , $\text{I}_2$ ) actually present in the reaction, and it treats coefficients as multipliers rather than exponents.

D

$K_c = \frac{2[\text{H}]2[\text{I}]}{[\text{HI}]^2}$

This is incorrect because it substitutes atomic concentrations for molecular products and uses coefficients as multipliers in the numerator.

Q3

2023

QCAA

Paper 1

1 mark

Q3

1 mark

Two 0.1 M acidic solutions, X and Y, are 100% dissociated. Solution X has an electrical conductivity approximately twice that of solution Y. Identify solutions X and Y.

	Solution X	Solution Y
A	HCl	CH $_3$ COOH
B	HNO $_3$	H $_2$ SO $_4$
C	H $_3$ PO $_4$	HNO $_3$
D	H $_2$ SO $_4$	HCl

A

Row A

B

Row B

C

Row C

D

Row D

Reveal Answer

A

Row A

Acetic acid (CH $_3$ COOH) is a weak acid and does not typically dissociate 100%. Even if hypothetically 100% dissociated, both HCl and CH $_3$ COOH are monoprotic acids producing equal concentrations of ions, resulting in similar conductivities rather than a 2:1 ratio.

B

Row B

This option reverses the required order. H $_2$ SO $_4$ (diprotic) would have a higher conductivity than HNO $_3$ (monoprotic), meaning Solution Y would be more conductive than Solution X, contradicting the problem statement.

C

Row C

H $_3$ PO $_4$ is a weak acid. If assumed to be 100% dissociated, it would release three H $^+$ ions, likely resulting in a conductivity ratio closer to 3:1 compared to the monoprotic HNO $_3$ .

D

Row D

Correct Answer

H $_2$ SO $_4$ is a diprotic acid yielding 2 moles of H $^+$ per mole of acid, while HCl is monoprotic yielding 1 mole of H $^+$ . Assuming 100% dissociation, Solution X (H $_2$ SO $_4$ ) will have twice the concentration of charge-carrying protons as Solution Y (HCl), resulting in approximately twice the conductivity.

Q3

2022

QCAA

Paper 2

9 marks

Q3

A 50.0 mL solution of ethanoic acid ( $CH_3COOH$ ) was titrated with 15.0 mL of 0.10 M sodium hydroxide (NaOH) solution to reach the equivalence point ( $pK_a$ ethanoic acid = 4.76).

Q3a

2 marks

Write a balanced chemical equation to indicate how ethanoic acid acts as a Brønsted-Lowry acid during the titration and identify its conjugate base.

Reveal Answer

$CH_3COOH(aq) + OH^-(aq) \rightleftharpoons CH_3COO^-(aq) + H_2O(l)$

The conjugate base formed is $CH_3COO^-$ .

Marking Criteria

Descriptor	Marks
Provides correct balanced chemical equation	1
Identifies $CH_3COO^-$ as conjugate base	1

Q3b

1 mark

Determine the $K_b$ of the conjugate base of ethanoic acid. (to two decimal places)

Reveal Answer

$K_b = \frac{1.00 \times 10^{-14}}{10^{-4.76}} = 10^{-9.24} = 5.75 \times 10^{-10}$

Marking Criteria

Descriptor	Marks
Determines $K_b$ is $5.75 \times 10^{-10}$	1

Q3c

2 marks

Calculate the concentration of the conjugate base at the equivalence point. Show your working. (to three significant figures)

Reveal Answer

At equivalence point, moles $OH^-$ = moles $H^+$ = moles $CH_3COO^-$
moles $CH_3COO^- = n \times V = 0.10 \times 0.015 = 1.50 \times 10^{-3}$

Marking Criteria

Descriptor	Marks
Determines moles $CH_3COO^-$ is $1.50 \times 10^{-3}$	1
Calculates $[CH_3COO^-]$ is $2.31 \times 10^{-2} mol L^{-1}$	1

Q3d

4 marks

Calculate the pH at the equivalence point. Show your working. (to one decimal place)

Reveal Answer

$[CH_3COOH] = [OH^-] = x$

$K_b = \frac{[OH^-][CH_3COOH]}{[CH_3COO^-]}$

$K_b = 5.75 \times 10^{-10} = \frac{x^2}{[2.31 \times 10^{-2}]}$

$x^2 = 1.33 \times 10^{-11}$

$x = 3.64 \times 10^{-6} = [OH^-]$

$pOH = -\log(3.64 \times 10^{-6}) = 5.44 \sim 5.4$

$pH = 14 - 5.4 = 8.6$

Marking Criteria

Descriptor	Marks
Provides correct substitution	1
Calculates $[OH^-]$ is $3.64 \times 10^{-6}$ M	1
Determines pOH is 5.4	1
Calculates pH is 8.6	1

Q28

2022

SCSA

7 marks

Q28

7 marks

Explain why potassium hydrogensulfite, $\mathrm{KHSO_3}$ , produces an acidic solution when dissolved in water, while potassium hydrogencarbonate, $\mathrm{KHCO_3}$ , produces a basic solution when dissolved in water. Use equations to illustrate your explanation.

Reveal Answer

The $K^+$ ions in solution are neutral and do not react with water, whereas the $HSO_3^-$ and $HCO_3^-$ ions undergo hydrolysis reactions. For the hydrolysis reactions for $HSO_3^-$ , the reaction that produces $H_3O^+$ occurs to a greater extent than the reaction that produces $OH^-$ , therefore the solution will be acidic. For the hydrolysis reactions for $HCO_3^-$ , the reaction that produces $OH^-$ occurs to a greater extent than the reaction that produces $H_3O^+$ , therefore the solution is basic. A basic solution has a greater concentration of $OH^-$ ions than $H_3O^+$ ions, and an acidic solution has a greater concentration of $H_3O^+$ ions than $OH^-$ ions.

The relevant equations for $HSO_3^-$ are (any 1 of the following):

$HSO_3^-(aq) + H_2O(l) \rightleftharpoons SO_3^{2-}(aq) + H_3O^+(aq)$
$HSO_3^-(aq) + H_2O(l) \rightleftharpoons H_2SO_3(aq) + OH^-(aq)$ .

The relevant equations for $HCO_3^-$ are (any 1 of the following):

$HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + OH^-(aq)$
$HCO_3^-(aq) + H_2O(l) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)$ .

Marking Criteria

Descriptor	Marks
Recognises that the $K^+$ ions in solution are neutral/do not react with water.	1
Recognises that the $HSO_3^-$ and $HCO_3^-$ ions undergo hydrolysis reactions.	1
Recognises that for the hydrolysis reactions for $HSO_3^-$ , the reaction that produces $H_3O^+$ occurs to a greater extent than the reaction that produces $OH^-$ , therefore the solution will be acidic.	1
Recognises that for the hydrolysis reactions for $HCO_3^-$ , the reaction that produces $OH^-$ occurs to a greater extent than the reaction that produces $H_3O^+$ , therefore the solution is basic.	1
Recognises that a basic solution has a greater concentration of $OH^-$ ions than $H_3O^+$ ions, or an acidic solution has a greater concentration of $H_3O^+$ ions than $OH^-$ ions.	1
Provides at least one appropriate equation for $HSO_3^-$ , such as $HSO_3^-(aq) + H_2O(l) \rightleftharpoons SO_3^{2-}(aq) + H_3O^+(aq)$ or $HSO_3^-(aq) + H_2O(l) \rightleftharpoons H_2SO_3(aq) + OH^-(aq)$ .	1
Provides at least one appropriate equation for $HCO_3^-$ , such as $HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + OH^-(aq)$ or $HCO_3^-(aq) + H_2O(l) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)$ .	1

Q14

2021

QCAA

Paper 1

1 mark

Q14

1 mark

Analyse the data to determine the relative strengths of acids from strongest to weakest.

Acid	$K_a$ value
Nitrous acid	$K_a = 4.00 \times 10^{-4}$
Ethanoic acid	$K_a = 1.76 \times 10^{-5}$
Hydrofluoric acid	$K_a = 7.20 \times 10^{-4}$
Chloroethanoic acid	$K_a = 1.40 \times 10^{-3}$

A

chloroethanoic, ethanoic, nitrous, hydrofluoric

B

chloroethanoic, hydrofluoric, nitrous, ethanoic

C

ethanoic, nitrous, hydrofluoric, chloroethanoic

D

ethanoic, hydrofluoric, nitrous, chloroethanoic

Reveal Answer

A

chloroethanoic, ethanoic, nitrous, hydrofluoric

This order is incorrect because ethanoic acid has the smallest $K_a$ value ( $1.76 \times 10^{-5}$ ), making it the weakest acid, so it should be listed last rather than second.

B

chloroethanoic, hydrofluoric, nitrous, ethanoic

Correct Answer

Acid strength corresponds to the magnitude of the $K_a$ value. The correct order from largest to smallest $K_a$ is chloroethanoic ( $1.40 \times 10^{-3}$ ) > hydrofluoric ( $7.20 \times 10^{-4}$ ) > nitrous ( $4.00 \times 10^{-4}$ ) > ethanoic ( $1.76 \times 10^{-5}$ ).

C

ethanoic, nitrous, hydrofluoric, chloroethanoic

This option ranks the acids from weakest to strongest (increasing $K_a$ ), but the question asks for the order from strongest to weakest.

D

ethanoic, hydrofluoric, nitrous, chloroethanoic

This option incorrectly lists ethanoic acid as the strongest; however, it has the lowest $K_a$ value ( $1.76 \times 10^{-5}$ ), which indicates it is actually the weakest acid in the group.

Q33

2023

SCSA

7 marks

Q33

A barium hydroxide solution is titrated against an ammonium chloride solution to produce barium chloride, ammonia and water.

Q33a

2 marks

Write a balanced ionic equation for this reaction.

Reveal Answer

$\text{OH}^-\text{(aq)} + \text{NH}_4^+\text{(aq)} \rightarrow \text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)}$

Marking Criteria

Descriptor	Marks
Correct species	1
Correct balancing	1

Q33b

5 marks

Consider the following indicators

Indicator	pH change range	Colour change
Methyl orange	3.1–4.4	Red to yellow
Bromothymol blue	6.2–7.6	Yellow to blue
Phenolphthalein	8.3–10.0	Colourless to pink

Identify the most appropriate indicator for this titration and justify your choice, using an equation to support your answer.

Reveal Answer

Phenolphthalein is the most appropriate indicator.

At the equivalence point, the solution is basic because ammonia is produced, and ammonia hydrolyses in water to form hydroxide ions:

$\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$

This means that at the equivalence point, $\mathrm{[OH^-] > [H^+]}$ , so the pH is greater than 7.

Therefore, the best indicator is phenolphthalein, because its colour change range is $\mathrm{pH\ 8.3 - 10.0}$ , which is in the basic range and matches the equivalence point of this titration.

Marking Criteria

Descriptor	Marks
Phenolphthalein	1
Recognition that (at the equivalence point) the ammonia hydrolyses to produce OH-	1
Recognition that at (equivalence point) [OH-]>[H+] (and solution is basic)	1
Recognition that indicator changes in the basic range/indicator colour change/end point is a similar pH to the equivalence point	1
Appropriate equation NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)	1

Q23

2025

NESA

3 marks

Q23

3 marks

A student attempted to determine the % w/w of sulfate in a sample of solid fertiliser. They used the procedure described below.

Weigh a clean, dry beaker.
Add fertiliser to the beaker and weigh again.
Add 250 mL of distilled water and stir thoroughly.
Add 20 mL of 0.1 mol L $^{-1}$ BaCl $_2$ solution.
Filter out the BaSO $_4$ precipitate, using distilled water to ensure all of the solid is transferred from the beaker to the filter paper.
Put the filter paper and precipitate onto a weighed watch glass and leave them to dry for 20 minutes in the sun.
Weigh the watch glass, the filter paper and the precipitate.
Calculate the % w/w.

Justify TWO changes that can be made to the procedure to ensure more accurate results.

Reveal Answer

Filter out any insoluble components before step 4, or else these will contribute to the mass of the precipitate.

Weigh the filter paper, or else the mass of the precipitate will be inaccurate.

Marking Criteria

Descriptor	Marks
Justifies TWO appropriate changes with justifications of both related to accuracy	3
Justifies ONE appropriate change OR Identifies TWO appropriate changes	2
Provides some relevant information	1
None of the above	0

Q1

2023

SCSA

1 mark

Q1

1 mark

Which of the following is likely to occur due to the increase of carbon dioxide levels in the atmosphere?

A

oceans cool and absorb less carbon dioxide from the atmosphere

B

it will be more difficult for crustations to construct their shells

C

the pH of oceans will increase, becoming more acidic

D

the availability of carbonate ions to marine organisms will increase

Reveal Answer

A

oceans cool and absorb less carbon dioxide from the atmosphere

Increased atmospheric carbon dioxide contributes to the greenhouse effect and global warming, which causes ocean temperatures to rise, not cool.

B

it will be more difficult for crustations to construct their shells

Correct Answer

Higher carbon dioxide levels lead to ocean acidification, which reduces the availability of carbonate ions that crustaceans and other marine organisms need to build their calcium carbonate shells.

C

the pH of oceans will increase, becoming more acidic

While the oceans will become more acidic due to increased carbon dioxide absorption, an increase in acidity corresponds to a decrease in pH, not an increase.

D

the availability of carbonate ions to marine organisms will increase

Ocean acidification increases the concentration of hydrogen ions, which react with carbonate ions to form bicarbonate, thereby decreasing the availability of carbonate ions for marine organisms.

Q4

2025

NESA

1 mark

Q4

1 mark

A student is presented with two clear colourless solutions. One contains Pb $^{2+}$ and the other Na $^+$ ions.

Which ion can be added to the solutions to identify the solutions?

A

I $^−$

B

NH $_4^+$

C

NO $_3^−$

D

CH $_3$ COO $^−$

Reveal Answer

A

I $^−$

Correct Answer

Adding iodide ions ( $I^-$ ) will form a distinct yellow precipitate of lead(II) iodide ( $PbI_2$ ) with $Pb^{2+}$ , while sodium iodide ( $NaI$ ) remains soluble, allowing the solutions to be easily distinguished.

B

NH $_4^+$

Ammonium ( $NH_4^+$ ) is a cation and will not react or form a precipitate with either $Pb^{2+}$ or $Na^+$ cations.

C

NO $_3^−$

All nitrate ( $NO_3^-$ ) salts are soluble in water, meaning no precipitate will form with either $Pb^{2+}$ or $Na^+$ to help distinguish them.

D

CH $_3$ COO $^−$

Most acetate ( $CH_3COO^-$ ) salts, including lead(II) acetate and sodium acetate, are soluble in water, so no visible reaction or precipitate will occur.

Q16

2020

SCSA

1 mark

Q16

1 mark

Which of the following is not a characteristic of a system in dynamic equilibrium?

A

The mass of the reactants equals the mass of the products.

B

Reactants are forming products and products are forming reactants.

C

The rates of the forward and reverse reactions are equal.

D

The position of the equilibrium is affected by temperature.

Reveal Answer

A

The mass of the reactants equals the mass of the products.

Correct Answer

This is the correct answer because it is not a characteristic of equilibrium. While the amounts of reactants and products remain constant at equilibrium, their masses or concentrations are rarely equal to each other.

B

Reactants are forming products and products are forming reactants.

This is a true characteristic of dynamic equilibrium. The system is "dynamic" precisely because both the forward and reverse reactions continue to occur.

C

The rates of the forward and reverse reactions are equal.

This is a true characteristic. The fundamental definition of dynamic equilibrium is that the forward and reverse reaction rates are exactly equal.

D

The position of the equilibrium is affected by temperature.

This is a true characteristic. According to Le Chatelier's principle, changing the temperature of a system at equilibrium will shift its position to favor either the endothermic or exothermic direction.

Q4

2025

SCSA

1 mark

Q4

1 mark

The equation for a system at equilibrium is given below.

$2 NO(g) + O_2(g) \leftrightharpoons 2 NO_2(g) + heat$

At 25 `C, the value of K for this equilibrium is $2.19 \times 10^{12}$ .

Which of the following statements about this system is true? Increasing the

A

partial pressure of NO(g) will increase the yield of $NO_2(g)$ and will increase the rate of the forward reaction.

B

partial pressure of NO(g) will increase the yield of $NO_2(g)$ but will decrease the rate of the forward reaction.

C

temperature will increase the yield of $NO_2(g)$ but decrease the rate of the forward reaction.

D

temperature will increase the yield of $NO_2(g)$ and increase the rate of the forward reaction.

Reveal Answer

A

partial pressure of NO(g) will increase the yield of $NO_2(g)$ and will increase the rate of the forward reaction.

Correct Answer

According to Le Chatelier's principle, increasing the partial pressure of a reactant ( $NO$ ) shifts the equilibrium toward the products, increasing the yield. Additionally, a higher concentration of reactants increases the frequency of collisions, thereby increasing the forward reaction rate.

B

partial pressure of NO(g) will increase the yield of $NO_2(g)$ but will decrease the rate of the forward reaction.

While increasing the partial pressure of a reactant does increase the yield, it also increases (rather than decreases) the reaction rate due to a higher frequency of molecular collisions.

C

temperature will increase the yield of $NO_2(g)$ but decrease the rate of the forward reaction.

Increasing temperature increases the kinetic energy of the molecules, which always leads to an increase in the reaction rate, not a decrease.

D

temperature will increase the yield of $NO_2(g)$ and increase the rate of the forward reaction.

The reaction forming $NO_2$ is exothermic; therefore, increasing the temperature shifts the equilibrium toward the reactants (left), decreasing the yield of $NO_2$ .

Q15

2025

VCAA

1 mark

Q15

1 mark

Consider the following two reactions that are at equilibrium at 500 °C.

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \qquad K_c = 2.86 \times 10^{-1}\ \mathrm{M^{-2}}$

$4NH_3(g) \rightleftharpoons 2N_2(g) + 6H_2(g)$

The magnitude of the value of $K_c$ for the second reaction is

A

$8.18 \times 10^{-2}$

B

$5.72 \times 10^{-1}$

C

$1.75 \times 10^0$

D

$1.22 \times 10^1$

Reveal Answer

A

$8.18 \times 10^{-2}$

This value is $(K_c)^2$ , which would be the equilibrium constant if the first reaction were multiplied by 2 but not reversed.

B

$5.72 \times 10^{-1}$

This value is $2 \times K_c$ . When a reaction is multiplied by a coefficient, the equilibrium constant must be raised to that power, not multiplied by it.

C

$1.75 \times 10^0$

This value results from incorrect mathematical manipulation of the equilibrium constant. The correct operation is to take the inverse square of the original $K_c$ .

D

$1.22 \times 10^1$

Correct Answer

The second reaction is the reverse of the first reaction multiplied by 2. Therefore, its equilibrium constant is $(K_c)^{-2} = (2.86 \times 10^{-1})^{-2} = 1.22 \times 10^1$ .

Q11

2024

QCAA

Paper 1

1 mark

Q11

1 mark

A Brønsted–Lowry acid

A

accepts a proton to form its base.

B

donates a proton to form its base.

C

accepts a proton to form its conjugate base.

D

donates a proton to form its conjugate base.

Reveal Answer

A

accepts a proton to form its base.

This describes a Brønsted–Lowry base, not an acid. Acids donate protons, whereas bases accept them.

B

donates a proton to form its base.

While acids do donate protons, the specific term for the species formed after the loss of a proton is the "conjugate base," making Option D the more precise and standard definition.

C

accepts a proton to form its conjugate base.

Acids donate protons rather than accepting them. Additionally, when a species accepts a proton, it forms a conjugate acid, not a conjugate base.

D

donates a proton to form its conjugate base.

Correct Answer

By definition, a Brønsted–Lowry acid is a proton ( $H^+$ ) donor. Once it donates a proton, the remaining species is called its conjugate base.

QCAA Chemistry Chemical equilibrium systems

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QCAA Chemistry Chemical equilibrium systems

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