NESA Chemistry Calculating the Equilibrium Constant
15 sample questions with marking guides and sample answers · Avg. score: 61.8%
The question refers to the decomposition of hydrogen iodide gas (HI) to produce hydrogen gas () and iodine gas () in a sealed 1-litre container.
Colourless Colourless Purple
Determine the equilibrium expression () for the reaction.
Reveal Answer
This is incorrect because the stoichiometric coefficient of the reactant (2) must be used as an exponent, not a multiplier. The correct term is , not .
This is correct based on the law of mass action for the reaction . The concentration of products is in the numerator, the reactant is in the denominator, and the coefficient 2 for HI becomes an exponent.
This is incorrect because it uses atomic species (H, I) instead of the molecular species (, ) actually present in the reaction, and it treats coefficients as multipliers rather than exponents.
This is incorrect because it substitutes atomic concentrations for molecular products and uses coefficients as multipliers in the numerator.
Consider the following two reactions that are at equilibrium at 500 °C.
The magnitude of the value of for the second reaction is
Reveal Answer
This value is , which would be the equilibrium constant if the first reaction were multiplied by 2 but not reversed.
This value is . When a reaction is multiplied by a coefficient, the equilibrium constant must be raised to that power, not multiplied by it.
This value results from incorrect mathematical manipulation of the equilibrium constant. The correct operation is to take the inverse square of the original .
The second reaction is the reverse of the first reaction multiplied by 2. Therefore, its equilibrium constant is .
Consider the following equilibrium system.
is pink and is blue. When a solution of these ions and chloride ions is heated, the mixture becomes more blue.
Relate the observed colour change to the change in .
Reveal Answer
The mixture becomes more blue as temperature is increased, which suggests that the concentration of is increasing. This means that an increase in temperature favours the forward reaction. This response will result in a larger value of as
| Descriptor | Marks |
|---|---|
Provides a sound relationship between the observed colour change and | 3 |
Provides some information about changes in concentration of reactants and/or products | 2 |
Provides some relevant information | 1 |
None of the above | 0 |
Which of the following statements about pure water are correct?
(i) Pure water is a weak electrolyte that undergoes self-ionisation.
(ii) The equilibrium constant for the ionisation of pure water at 25 °C is .
(iii) Pure water ionises completely at 25 °C, hence [H⁺] = [OH⁻].
(iv) The ionisation of pure water produces twice as many hydrogen ions as hydroxide ions.
i and ii only
ii and iii only
iii and iv only
i, ii, iii and iv
Reveal Answer
i and ii only
Statements (i) and (ii) are correct. Pure water is a weak electrolyte that slightly self-ionises, and its ion-product constant () at 25 °C is .
ii and iii only
Statement (iii) is incorrect. While it is true that in pure water, water only ionises to a very small extent, not completely.
iii and iv only
Statements (iii) and (iv) are both incorrect. Water does not ionise completely, and its self-ionisation produces equal amounts of hydrogen and hydroxide ions, not twice as many.
i, ii, iii and iv
This option is incorrect because statements (iii) and (iv) are false. Water only partially ionises and produces a 1:1 ratio of hydrogen to hydroxide ions.
Two experiments were conducted to investigate the effect of temperature on the equilibrium formed during the decomposition of hydrogen iodide (HI).
| Experiment | Initial [HI] | Initial [H] | Initial [I] | Eq [HI] | Eq [H] | Eq [I] | |
|---|---|---|---|---|---|---|---|
| 1 | 0.08 | 0.00 | 0.00 | 0.01 | 2.78 10 | ||
| 2 | 0.00 | 0.06 | 0.06 | 0.06 | 0.03 | 0.03 |
(Note: Concentrations are in mol L)
Determine the concentration of HI(g) and I(g) at equilibrium for experiment 1.
[HI]:
[I]:
Reveal Answer
| Descriptor | Marks |
|---|---|
determines [HI] is 0.06 | 1 |
determines [I₂] is 0.01 | 1 |
Calculate the equilibrium constant () for experiment 2. Show your working.
Reveal Answer
| Descriptor | Marks |
|---|---|
uses appropriate substitution | 1 |
calculates is 0.25 | 1 |
Determine which experiment was conducted at a higher temperature. Explain your reasoning.
Reveal Answer
Temperature was higher for experiment 2. The value was larger for experiment 2, indicating that the equilibrium shifted towards the products (endothermic direction) to compensate for the increase in temperature.
| Descriptor | Marks |
|---|---|
determines experiment 2 was conducted at a higher temperature | 1 |
explains that value for experiment 2 is larger | 1 |
explains higher value indicates equilibrium shifts to the right | 1 |
Consider the following reversible reaction:
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?
The concentrations of reactants and products will be equal.
There will be a greater concentration of products than reactants.
The reactant concentration will be greater than that of the products.
The concentrations of NOCl and NO will be double the concentration of .
Reveal Answer
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 is significantly different from 1.
There will be a greater concentration of products than reactants.
Because the equilibrium constant is much greater than 1 (), the equilibrium lies far to the right, meaning the formation of products is strongly favored.
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 was much less than 1, indicating a reactant-favored system.
The concentrations of NOCl and NO will be double the concentration of .
The coefficients in the balanced chemical equation dictate the stoichiometric ratio in which substances react and form, not their absolute concentrations at equilibrium.
This table shows the effect of temperature on the pH of pure water.
| Temperature (°C) | pH |
|---|---|
| 10 | 7.27 |
| 15 | 7.17 |
| 20 | 7.08 |
| 25 | 7.00 |
| 30 | 6.92 |
| 50 | 6.63 |
Analyse the data to explain whether the self-ionisation of water is endothermic or exothermic. Explain your reasoning.
Reveal Answer
As the temperature increases, the [HO] increases.
2HO(l) HO(aq) + OH(aq)
Therefore, the equilibrium shifts towards the products.
Increasing temperature shifts equilibrium in the endothermic direction, therefore the self-ionisation of water is endothermic.
| Descriptor | Marks |
|---|---|
Identifies [HO] increases as temperature increases | 1 |
Identifies equilibrium shifts towards the products and the endothermic direction | 1 |
Determines self-ionisation of water is endothermic | 1 |
Calculate the of pure water at 50 °C. Show your working. (to three significant figures)
Reveal Answer
pH =
(to 3 significant figures)
| Descriptor | Marks |
|---|---|
Determines | 1 |
Determines consequentially correct | 1 |
Determine the expression for the weak base shown in the equilibrium equation.
Reveal Answer
This expression incorrectly places the reactants in the numerator and the products in the denominator. Additionally, pure liquids like are excluded from equilibrium expressions.
This option incorrectly places reactants in the numerator and includes liquid water. Equilibrium expressions follow the ratio .
The expression is the product of the concentrations of the products divided by the reactant concentration (). Pure liquids, such as , are omitted from the expression.
This expression incorrectly includes water in the denominator. Pure liquids are omitted from equilibrium constants, and the reactant is missing.
At 327 °C, the equilibrium constant for the reaction is .
What is the equilibrium constant at 327 °C for the reaction ?
Reveal Answer
Incorrect. When a reaction equation is multiplied by a factor of 2, the equilibrium constant must be squared, not multiplied by 2.
Incorrect. Although the units are correctly squared, the numerical value was incorrectly multiplied by 2 instead of being squared.
Incorrect. The numerical value was correctly squared, but the units must also be squared from to .
Correct. When a reaction equation is multiplied by a factor of 2, the new equilibrium constant is the square of the original. Therefore, .
Water and octan-1-ol do not mix. When an aqueous solution of bromoacetic acid () is shaken with octan-1-ol, an equilibrium system is established between bromoacetic acid dissolved in the octan-1-ol and in the water.
The equilibrium constant expression for this system is
An aqueous solution of bromoacetic acid with an initial concentration of is shaken with an equal volume of octan-1-ol. Bromoacetic acid does not dissociate in octan-1-ol but does dissociate in water, with . When the system has reached equilibrium, the is .
Calculate the equilibrium concentration of aqueous bromoacetic acid and hence, or otherwise, calculate the for the octan-1-ol and water system.
Reveal Answer
| Descriptor | Marks |
|---|---|
Calculates the | 4 |
Provides most steps of the calculation | 3 |
Provides some steps of the calculation | 2 |
Provides some relevant information | 1 |
None of the above | 0 |
In the conversion between sulfur dioxide and sulfur trioxide in a sealed vessel, the following equilibrium is established:
For this system, which of the following statements about the equilibrium constant K is correct? K will
increase if the temperature of the system is decreased.
decrease if the partial pressure of is increased.
increase if the temperature of the system is increased.
increase if the pressure of the system is increased.
Reveal Answer
increase if the temperature of the system is decreased.
The reaction is endothermic (). Decreasing the temperature shifts the equilibrium to the left, which decreases the equilibrium constant .
decrease if the partial pressure of is increased.
Changing the partial pressure of a reactant or product shifts the equilibrium position, but it does not change the value of the equilibrium constant , which is only temperature-dependent.
increase if the temperature of the system is increased.
Because the reaction is endothermic (), increasing the temperature shifts the equilibrium to the right to absorb the added heat, resulting in a higher equilibrium constant .
increase if the pressure of the system is increased.
Changing the total pressure of the system will shift the equilibrium position (in this case, to the left where there are fewer moles of gas), but it will not change the equilibrium constant .
Which of the following gives the equilibrium constant expression for the following equilibrium?
Reveal Answer
This is the correct expression. It follows the law of mass action by placing product concentrations in the numerator and reactant concentrations in the denominator, raising them to the power of their coefficients, and omitting the solid .
This expression is inverted. The equilibrium constant is defined as the ratio of product concentrations to reactant concentrations, not the other way around.
This option is both inverted (reactants over products) and incorrectly includes the solid . Pure solids are excluded from equilibrium expressions because their concentration remains constant.
This option incorrectly includes the concentration of the solid reactant, . In heterogeneous equilibria, pure solids and liquids are omitted from the expression.
Ammonia gas reacts with oxygen gas in the following equilibrium reaction.
The equilibrium expression for the reaction is
Reveal Answer
This option incorrectly places reactants in the numerator and products in the denominator, and it fails to raise the concentrations to the power of their stoichiometric coefficients.
While this option correctly places products over reactants, it fails to raise the concentrations to the power of their stoichiometric coefficients found in the balanced equation.
This expression is the inverse of the correct equilibrium constant; it incorrectly places the reactants in the numerator and the products in the denominator.
The equilibrium expression is calculated as the product of product concentrations divided by the product of reactant concentrations, with each raised to the power of its stoichiometric coefficient: .
The thermal decomposition of lithium peroxide () is given by the equation shown.
Mixtures of , and were allowed to reach equilibrium in two identical, closed containers, P and Q, at the same temperature. The amount of in container P is double that in container Q. The amount of is the same in each container.
What is the ratio of in container P to in container Q?
1 : 1
2 : 1
3 : 2
5 : 4
Reveal Answer
1 : 1
The equilibrium constant expression is because solids are excluded. Since the temperature is the same in both containers, the equilibrium concentration of is identical, making the ratio 1:1.
2 : 1
This incorrectly assumes the concentration of is proportional to the amount of solid . The amounts of pure solids do not affect the equilibrium position.
3 : 2
This ratio has no basis in the equilibrium expression. The amounts of solid reactants and products do not change the equilibrium concentration of the gas.
5 : 4
This is incorrect because the equilibrium concentration of depends only on the temperature-dependent equilibrium constant, not on the initial amounts of solids.
The equilibrium concentration of A is M and B is M.
Which option represents the ratio of molecules present in a sample of the gaseous mixture when the temperature is decreased and a new equilibrium established?
8 molecules of A and 2 molecules of B
5 molecules of A and 5 molecules of B
3 molecules of A and 7 molecules of B
2 molecules of A and 8 molecules of B
Reveal Answer
8 molecules of A and 2 molecules of B
The reaction is endothermic (), so decreasing the temperature shifts the equilibrium to the left (reactants) to generate heat. Since the initial concentration of A was already higher than B (), shifting left increases the ratio of A to B further, making the 8:2 distribution the correct representation.
5 molecules of A and 5 molecules of B
This option represents a 1:1 ratio of A to B. Since the initial mixture already contained more A than B and the equilibrium shifts further toward A upon cooling, the final mixture must have a ratio of A to B significantly greater than 1.
3 molecules of A and 7 molecules of B
This option shows a higher amount of product B than reactant A. Decreasing the temperature of an endothermic reaction favors the reverse reaction, which would decrease the amount of B and increase the amount of A.
2 molecules of A and 8 molecules of B
This option implies a shift toward the product B. According to Le Chatelier's principle, removing heat (lowering temperature) from an endothermic reaction drives the system toward the reactants, so A should be the predominant species.