SCSA Chemistry Chemical equilibrium systems

15 sample questions with marking guides and sample answers

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.

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)+O2(g)SO2(g)+297kJS(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 V2O5V_2O_5 catalyst at a pressure of between 100 and 200 kPa:

2SO2(g)+O2(g)2SO3(g)+198kJ2 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 (H2S2O7H_2S_2O_7). Water is added to the oleum, producing 18 mol L1^{-1} sulfuric acid:

SO3(g)+H2SO4(l)H2S2O7(l)SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l) H2S2O7(l)+H2O(l)2H2SO4(aq)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 SO3_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 SO3_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

DescriptorMarks

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

DescriptorMarks

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

DescriptorMarks

States that high pressures are costly (and dangerous).

1

Compromise

DescriptorMarks

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
Q15
2021
SCSA
1 mark
Q15
1 mark

Which of the following processes does not contribute to the building of weaker seashells through ocean acidification?

A

HCO3(aq)+H2O()CO32(aq)+H3O+(aq)HCO_3^-(aq) + H_2O(\ell) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)

B

2H+(aq)+CaCO3(s)Ca2+(aq)+CO2(aq)+H2O()2 H^+(aq) + CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_2(aq) + H_2O(\ell)

C

CO2(g)+H2O()H2CO3(aq)CO_2(g) + H_2O(\ell) \rightleftharpoons H_2CO_3(aq)

D

H2CO3(aq)+H2O()HCO3(aq)+H3O+(aq)H_2CO_3(aq) + H_2O(\ell) \rightleftharpoons HCO_3^-(aq) + H_3O^+(aq)

Reveal Answer
A

HCO3(aq)+H2O()CO32(aq)+H3O+(aq)HCO_3^-(aq) + H_2O(\ell) \rightleftharpoons CO_3^{2-}(aq) + H_3O^+(aq)

This equilibrium determines the concentration of carbonate ions (CO32CO_3^{2-}). Increased ocean acidity shifts this reaction to the left, reducing the carbonate available for marine organisms to build strong shells.

B

2H+(aq)+CaCO3(s)Ca2+(aq)+CO2(aq)+H2O()2 H^+(aq) + CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_2(aq) + H_2O(\ell)

Correct Answer

This reaction represents the dissolution of already-formed calcium carbonate (CaCO3CaCO_3) shells by increased acidity. While it destroys existing shells, it is not a process that inhibits the initial building of new shells.

C

CO2(g)+H2O()H2CO3(aq)CO_2(g) + H_2O(\ell) \rightleftharpoons H_2CO_3(aq)

This reaction shows carbon dioxide dissolving in water to form carbonic acid, which is the primary driver of ocean acidification that ultimately impairs shell building.

D

H2CO3(aq)+H2O()HCO3(aq)+H3O+(aq)H_2CO_3(aq) + H_2O(\ell) \rightleftharpoons HCO_3^-(aq) + H_3O^+(aq)

This reaction shows the dissociation of carbonic acid into bicarbonate and hydronium ions, which increases ocean acidity and leads to a reduction in the carbonate ions needed for shell building.

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 (H2\text{H}_2) and iodine gas (I2\text{I}_2) in a sealed 1-litre container.

2HI(g)H2(g)+I2(g)ΔH=+53.6 kJ mol12\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 (KcK_c) for the reaction.

A

Kc=[H2][I2]2[HI]K_c = \frac{[\text{H}_2][\text{I}_2]}{2[\text{HI}]}

B

Kc=[H2][I2][HI]2K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

C

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

D

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

Reveal Answer
A

Kc=[H2][I2]2[HI]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 [HI]2[\text{HI}]^2, not 2[HI]2[\text{HI}].

B

Kc=[H2][I2][HI]2K_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 2HIH2+I22\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

Kc=2[H]2[I]2[HI]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 (H2\text{H}_2, I2\text{I}_2) actually present in the reaction, and it treats coefficients as multipliers rather than exponents.

D

Kc=2[H]2[I][HI]2K_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.

Q7
2023
VCAA
1 mark
Q7
1 mark

Consider the following statements about fossil fuels and biofuels.

I. Production of biofuels does not damage the environment.
II. Combustion of both biofuels and fossil fuels generates greenhouse gases.
III. Biofuels and fossil fuels are both renewable as they are produced from plants.

Which of the statements above are correct?

A

I only

B

II only

C

I and II only

D

I and III only

Reveal Answer
A

I only

This option is incorrect because Statement I is false. The production of biofuels can damage the environment through deforestation, habitat destruction, and the use of agricultural fertilizers.

B

II only

Correct Answer

This is the correct option because only Statement II is true. The combustion of both biofuels and fossil fuels releases carbon dioxide (CO2CO_2), which is a greenhouse gas.

C

I and II only

This option is incorrect because Statement I is false. While Statement II is true, biofuel production is not entirely harmless to the environment.

D

I and III only

This option is incorrect because both Statements I and III are false. Fossil fuels are considered non-renewable because they take millions of years to form, unlike biofuels which can be replenished relatively quickly.

Q15
2025
VCAA
1 mark
Q15
1 mark

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

N2(g)+3H2(g)2NH3(g)Kc=2.86×101 M2N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \qquad K_c = 2.86 \times 10^{-1}\ \mathrm{M^{-2}}

4NH3(g)2N2(g)+6H2(g)4NH_3(g) \rightleftharpoons 2N_2(g) + 6H_2(g)

The magnitude of the value of KcK_c for the second reaction is

A

8.18×1028.18 \times 10^{-2}

B

5.72×1015.72 \times 10^{-1}

C

1.75×1001.75 \times 10^0

D

1.22×1011.22 \times 10^1

Reveal Answer
A

8.18×1028.18 \times 10^{-2}

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

B

5.72×1015.72 \times 10^{-1}

This value is 2×Kc2 \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

1.75×1001.75 \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 KcK_c.

D

1.22×1011.22 \times 10^1

Correct Answer

The second reaction is the reverse of the first reaction multiplied by 2. Therefore, its equilibrium constant is (Kc)2=(2.86×101)2=1.22×101(K_c)^{-2} = (2.86 \times 10^{-1})^{-2} = 1.22 \times 10^1.

Q9
2023
VCAA
1 mark
Q9
1 mark

Consider the following statements about coal seam gas and petroleum gas.

I. Coal seam gas and petroleum gas are both mixtures.
II. Coal seam gas and petroleum gas both combust to produce carbon dioxide.
III. Coal seam gas and petroleum gas are both fossil fuels.

Which of the above statements are correct?

A

I and II only

B

I and III only

C

II and III only

D

I, II and III

Reveal Answer
A

I and II only

While statements I and II are correct, statement III is also true because both coal seam gas and petroleum gas are formed from ancient organic matter, making them fossil fuels.

B

I and III only

While statements I and III are correct, statement II is also true because both gases are primarily composed of hydrocarbons, which produce carbon dioxide and water upon combustion.

C

II and III only

While statements II and III are correct, statement I is also true because both gases are mixtures of various hydrocarbons and other trace gases, rather than pure substances.

D

I, II and III

Correct Answer

All three statements are correct. Both gases are mixtures of hydrocarbons, combust to produce carbon dioxide, and are classified as fossil fuels.

Q32
2024
NESA
4 marks
Q32
4 marks

Calculate the concentration of cadmium ions in a saturated solution of cadmium(II) phosphate, Cd3(PO4)2\text{Cd}_3(\text{PO}_4)_2, Ksp=2.53×1033K_{sp} = 2.53 \times 10^{-33}.

Reveal Answer

Cd3(PO4)2(s)3Cd2+(aq)+2PO43(aq)\text{Cd}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Cd}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)

Let s mol Cd3(PO4)2(s)s \text{ mol } \text{Cd}_3(\text{PO}_4)_2(s) dissolve per litre

Ksp=[Cd2+]3[PO43]2=(3s)3(2s)2=108s5K_{sp} = [\text{Cd}^{2+}]^3[\text{PO}_4^{3-}]^2 = (3s)^3(2s)^2 = 108s^5

s=2.53×10331085=1.19×107s = \sqrt[5]{\frac{2.53 \times 10^{-33}}{108}} = 1.19 \times 10^{-7}

[Cd2+]=3s=3.56×107 mol L1[\text{Cd}^{2+}] = 3s = 3.56 \times 10^{-7} \text{ mol L}^{-1}

Marking Criteria
DescriptorMarks

Calculates the concentration of Cd2+\text{Cd}^{2+} ions to 3 significant figures

4

Provides most relevant steps of the calculation

3

Provides some relevant steps of the calculation

2

Provides some relevant information

1

None of the above

0
Q4
2025
SCSA
1 mark
Q4
1 mark

The equation for a system at equilibrium is given below.

2NO(g)+O2(g)2NO2(g)+heat2 NO(g) + O_2(g) \leftrightharpoons 2 NO_2(g) + heat

At 25 `C, the value of K for this equilibrium is 2.19×10122.19 \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 NO2(g)NO_2(g) and will increase the rate of the forward reaction.

B

partial pressure of NO(g) will increase the yield of NO2(g)NO_2(g) but will decrease the rate of the forward reaction.

C

temperature will increase the yield of NO2(g)NO_2(g) but decrease the rate of the forward reaction.

D

temperature will increase the yield of NO2(g)NO_2(g) and increase the rate of the forward reaction.

Reveal Answer
A

partial pressure of NO(g) will increase the yield of NO2(g)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 (NONO) 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 NO2(g)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 NO2(g)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 NO2(g)NO_2(g) and increase the rate of the forward reaction.

The reaction forming NO2NO_2 is exothermic; therefore, increasing the temperature shifts the equilibrium toward the reactants (left), decreasing the yield of NO2NO_2.

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.

Q2
2022
QCAA
Paper 2
8 marks
Q2

The reaction shows part of the contact process used to produce sulfuric acid.

2SO2(g)+O22SO3(g)2SO_2(g) + O_2 \rightleftharpoons 2SO_3(g)

The equilibrium constant (KcK_c) for this reaction at different temperatures is shown.

Temperature (K)Equilibrium constant, KcK_c (mol L1L^{-1})
2989.77×10259.77 \times 10^{25}
5008.61×10118.61 \times 10^{11}
Q2a
2 marks

Deduce if the forward reaction is exothermic or endothermic. Explain your reasoning.

Reveal Answer

Exothermic
Increasing temperature decreases KcK_c, indicating that the equilibrium shifts towards the reactants (endothermic) direction.

Marking Criteria
DescriptorMarks

Determines forward reaction is exothermic

1

Explains that the decrease in KcK_c as temperature increases indicates endothermic direction is towards the reactants

1
Q2b
2 marks

Calculate the equilibrium concentration of SO3SO_3 at 500 K given the equilibrium concentrations.

[SO2]=0.860 M;[O2]=0.330 M[SO_2] = 0.860\ \text{M}; [O_2] = 0.330\ \text{M}
Concentration = _____ M (to three significant figures)

Reveal Answer

8.61×1011=[SO3]2(0.860)2(0.330)8.61 \times 10^{11} = \frac{[SO_3]^2}{(0.860)^2(0.330)}

[SO3]2=(0.7396)(0.330)(8.61×1011)=2.10×1011[SO_3]^2 = (0.7396)(0.330)(8.61 \times 10^{11}) = 2.10 \times 10^{11}

Concentration = 4.58×1054.58 \times 10^5 M

Marking Criteria
DescriptorMarks

Provides appropriate working

1

Calculates [SO3]=4.58×105[SO_3] = 4.58 \times 10^5

1
Q2c
4 marks

Apply Le Châtelier’s principle to explain whether halving the reaction vessel’s volume at 500 K would affect the position of the equilibrium or the value of the equilibrium constant.

Reveal Answer

Halving the volume would double the pressure.
To reduce the pressure, the equilibrium would shift toward the product to reduce the number of molecules present.
However, due to the reactant decreasing the equilibrium constant remains unchanged.

Marking Criteria
DescriptorMarks

Indicates that halving the volume doubles the pressure

1

Explains that equilibrium will shift to reduce the number of molecules present to reduce pressure

1

Explains that equilibrium will shift toward the product

1

Explains why the equilibrium constant would not change

1
Q11
2024
NESA
1 mark
Q11
1 mark

Which is the correct expression for calculating the solubility (in mol L1\text{mol L}^{-1}) of lead(II) iodide in a 0.1 mol L10.1 \text{ mol L}^{-1} solution of NaI at 25C25^\circ\text{C}?

A
9.8×1092×0.1\frac{9.8 \times 10^{-9}}{2 \times 0.1}
B
9.8×109(2×0.1)2\frac{9.8 \times 10^{-9}}{(2 \times 0.1)^2}
C
9.8×1090.1\frac{9.8 \times 10^{-9}}{0.1}
D
9.8×109(0.1)2\frac{9.8 \times 10^{-9}}{(0.1)^2}
Reveal Answer
A
9.8×1092×0.1\frac{9.8 \times 10^{-9}}{2 \times 0.1}

This option incorrectly assumes the iodide concentration is 2×0.12 \times 0.1 and fails to square the iodide concentration in the KspK_{sp} expression.

B
9.8×109(2×0.1)2\frac{9.8 \times 10^{-9}}{(2 \times 0.1)^2}

This option incorrectly assumes the iodide concentration from NaI is 2×0.12 \times 0.1. NaI only produces one iodide ion per formula unit, so the initial [I][\text{I}^-] is 0.1 mol L10.1 \text{ mol L}^{-1}.

C
9.8×1090.1\frac{9.8 \times 10^{-9}}{0.1}

This option forgets to square the iodide concentration. The KspK_{sp} expression for PbI2\text{PbI}_2 is [Pb2+][I]2[\text{Pb}^{2+}][\text{I}^-]^2, so the denominator must be squared.

D
9.8×109(0.1)2\frac{9.8 \times 10^{-9}}{(0.1)^2}
Correct Answer

The KspK_{sp} expression is [Pb2+][I]2[\text{Pb}^{2+}][\text{I}^-]^2. In a 0.1 mol L10.1 \text{ mol L}^{-1} NaI solution, [I]0.1 mol L1[\text{I}^-] \approx 0.1 \text{ mol L}^{-1}. Therefore, the solubility s=[Pb2+]=Ksp(0.1)2s = [\text{Pb}^{2+}] = \frac{K_{sp}}{(0.1)^2}.

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.

Q12
2024
VCAA
1 mark
Q12
1 mark

The combustion reaction between butane gas, C4H10\text{C}_4\text{H}_{10}, and oxygen gas, O2\text{O}_2, is considered irreversible because

A

the forward reaction is exothermic.

B

the products are less stable than the reactants.

C

the rate of the reverse reaction is so slow that it can be ignored.

D

an unlimited supply of oxygen will favour the forward reaction.

Reveal Answer
A

the forward reaction is exothermic.

While combustion is highly exothermic, exothermicity alone does not make a reaction irreversible. Many reversible reactions have exothermic forward reactions.

B

the products are less stable than the reactants.

In a combustion reaction, the products (CO2\text{CO}_2 and H2O\text{H}_2\text{O}) are actually more stable than the reactants, which is why the reaction releases energy.

C

the rate of the reverse reaction is so slow that it can be ignored.

Correct Answer

A reaction is considered practically irreversible when the activation energy for the reverse reaction is so high that its rate is effectively zero. This means the products do not convert back into reactants under normal conditions.

D

an unlimited supply of oxygen will favour the forward reaction.

While adding a reactant like oxygen favors the forward reaction according to Le Chatelier's principle, this describes shifting an equilibrium rather than defining a reaction as irreversible.

Q20
2021
SCSA
1 mark
Q20
1 mark

Consider the following reversible reaction:

2NO(g)+Cl2(g)2NOCl(g)Kc=6.5×104 at 35 °C2 NO(g) + Cl_2(g) \rightleftharpoons 2 NOCl(g) \qquad K_c = 6.5 \times 10^4 \text{ at 35 °C}

Which of the following statements describes the relative concentrations of reactants and products in this system when equilibrium is established in a closed vessel at 35 °C?

A

The concentrations of reactants and products will be equal.

B

There will be a greater concentration of products than reactants.

C

The reactant concentration will be greater than that of the products.

D

The concentrations of NOCl and NO will be double the concentration of Cl2Cl_2.

Reveal Answer
A

The concentrations of reactants and products will be equal.

Equal concentrations of reactants and products do not generally occur at equilibrium, especially when the equilibrium constant KcK_c is significantly different from 1.

B

There will be a greater concentration of products than reactants.

Correct Answer

Because the equilibrium constant KcK_c is much greater than 1 (6.5×1046.5 \times 10^4), the equilibrium lies far to the right, meaning the formation of products is strongly favored.

C

The reactant concentration will be greater than that of the products.

A greater concentration of reactants than products would only be expected if the equilibrium constant KcK_c was much less than 1, indicating a reactant-favored system.

D

The concentrations of NOCl and NO will be double the concentration of Cl2Cl_2.

The coefficients in the balanced chemical equation dictate the stoichiometric ratio in which substances react and form, not their absolute concentrations at equilibrium.

Frequently Asked Questions

How many SCSA Chemistry questions cover Chemical equilibrium systems?
AusGrader has 226 SCSA Chemistry questions on Chemical equilibrium systems, all with instant AI grading and detailed marking feedback.

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