ESAT Chemistry: Acids, rates and energetics, set 3

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   In a titration, 20.0 cm3 of 0.30 mol/dm3 sodium hydroxide solution exactly neutralises 30.0 cm3 of dilute hydrochloric acid. Sodium hydroxide and hydrochloric acid react in a 1:1 mole ratio.

   What is the concentration of the hydrochloric acid?

   1. A 0.45 mol/dm3
   2. B 0.12 mol/dm3
   3. C 0.2 mol/dm3
   4. D 0.1 mol/dm3
2. 21 mark

   A concentrated acid has pH 1. It is diluted with water until its H+ ion concentration has fallen to one thousandth (1/1000) of its original value.

   What is the pH of the diluted solution?

   1. A pH 4
   2. B pH 1000
   3. C pH 3
   4. D pH -2
3. 31 mark

   Aluminium oxide, Al2O3, is described as an amphoteric oxide.

   Which statement correctly describes its chemical behaviour?

   1. A It reacts only with acids, behaving as a typical basic oxide, and does not react with alkalis.
   2. B It reacts with neither acids nor alkalis, since 'amphoteric' means the oxide is chemically unreactive.
   3. C It reacts only with alkalis, behaving as a typical acidic oxide, and does not react with acids.
   4. D It reacts with both acids and alkalis, since an amphoteric oxide can behave as either a base or an acid depending on what it reacts with.
4. 41 mark

   Copper(II) oxide is insoluble in water. A student claims that copper(II) oxide cannot be a base, because it does not dissolve in water to form an alkaline solution.

   Is the student correct?

   1. A Yes, because a base must be able to dissolve in water to release OH-(aq) ions, and an insoluble oxide such as copper(II) oxide cannot do this.
   2. B No, because a base is more broadly defined as a substance that can accept H+ ions (or form OH-(aq) ions), and copper(II) oxide does accept H+ ions from an acid, even though it does not dissolve in water to do so.
   3. C No, because all metal oxides, whether or not they dissolve in water, are properly described as alkalis.
   4. D Yes, because only the oxides of Group 1 and Group 2 metals can ever behave as bases, and copper is in neither of these groups.
5. 51 mark

   A student mixes dilute nitric acid with potassium hydroxide solution in an insulated cup and measures a temperature rise from 18.0 degC to 25.0 degC.

   Which feature of an energy level diagram for this reaction is consistent with this observation?

   1. A The products of the reaction are drawn at a lower position on the energy level diagram than the reactants, since the exothermic neutralisation has released energy to the surroundings, raising their temperature.
   2. B The products are drawn at a higher position than the reactants, since additional energy has been absorbed by the reaction mixture in order to raise its own temperature.
   3. C The activation energy of this reaction must be zero, since a temperature rise was detectable as soon as the two solutions were mixed.
   4. D The reactants and products must be drawn at the same energy level, since this neutralisation reaction reaches a chemical equilibrium.
6. 61 mark

   Magnesium ribbon reacts with dilute sulfuric acid to produce magnesium sulfate solution and hydrogen gas.

   Which one of these changes would NOT increase the rate of this reaction?

   1. A Increasing the temperature of the acid.
   2. B Increasing the concentration of the sulfuric acid.
   3. C Cutting the magnesium ribbon into smaller pieces to increase its surface area.
   4. D Increasing the pressure of the surroundings around the reaction vessel.
7. 71 mark

   Sodium thiosulfate solution is mixed with dilute hydrochloric acid. The reaction produces a fine, insoluble precipitate of sulfur, which gradually makes the mixture cloudy and opaque; a trace of sulfur dioxide gas also escapes, but in too small and hazardous a quantity to be conveniently collected.

   Which method is most appropriate for measuring the rate of this reaction?

   1. A Timing how long it takes for a cross drawn on paper beneath the flask to become obscured by the increasing cloudiness of the reaction mixture.
   2. B Collecting the sulfur dioxide gas produced in a gas syringe and reading its volume at regular time intervals.
   3. C Using a colorimeter to measure the absorbance of the reaction mixture at regular time intervals, since the mixture develops colour as the reaction proceeds.
   4. D Weighing the sealed reaction flask at regular time intervals, since a gas is produced as the reaction proceeds.
8. 81 mark

   In an experiment measuring the volume of gas produced over time, 10 cm3 of gas had been collected after 10 seconds, and 40 cm3 had been collected after 25 seconds.

   Calculate the mean rate of reaction between 10 and 25 seconds.

   1. A 1.6 cm3/s
   2. B 0.5 cm3/s
   3. C 2 cm3/s
   4. D 1.2 cm3/s
9. 91 mark

   A fixed mass of magnesium ribbon reacts with excess dilute hydrochloric acid. The experiment is repeated with double the concentration of acid, at the same temperature.

   Which statement correctly uses collision theory to explain the resulting change in rate?

   1. A Doubling the concentration increases the frequency of collisions between acid particles and the magnesium surface, since more acid particles occupy the same volume of solution; the proportion of collisions with sufficient energy to react is unchanged.
   2. B Doubling the concentration increases the average kinetic energy of the acid particles, so a greater proportion of collisions now exceed the activation energy.
   3. C Doubling the concentration lowers the activation energy required for a successful collision between an acid particle and the magnesium surface.
   4. D Doubling the concentration has no effect on the rate, since the mass of magnesium and the temperature of the experiment have not changed.
10. 101 mark

    For an uncatalysed reaction, the reactants are at 80 kJ/mol and the peak of the energy level diagram is at 250 kJ/mol.

    Adding a catalyst lowers the peak to 190 kJ/mol, without changing the energy of the reactants or the products. By how much has the activation energy decreased as a result of the catalyst?

    1. A 170 kJ/mol
    2. B 110 kJ/mol
    3. C 190 kJ/mol
    4. D 60 kJ/mol
11. 111 mark

    A reversible reaction is allowed to reach equilibrium, once with a catalyst present and once without, at the same temperature.

    Which statement correctly compares the two cases?

    1. A The catalyst increases the equilibrium yield of product, since it speeds up the forward reaction more than the reverse reaction.
    2. B The catalyst causes equilibrium to be reached in a shorter time, but the equilibrium yield of product is unchanged, since it speeds up the forward and reverse reactions by the same amount.
    3. C The catalyst decreases the equilibrium yield of product, since it favours the reverse reaction over the forward reaction.
    4. D The catalyst has no effect at all, neither on how quickly equilibrium is reached nor on the equilibrium yield.
12. 121 mark

    Substance X can be converted directly into substance Z, with an enthalpy change of delta H1.

    The same overall conversion can also happen in two steps, via an intermediate substance Y: the step X -> Y has an enthalpy change of delta H2 = +130 kJ/mol, and the step Y -> Z has an enthalpy change of delta H3 = -210 kJ/mol.

    Use Hess's Law to calculate delta H1, the enthalpy change for the direct conversion of X into Z.

    1. A +340 kJ/mol
    2. B -210 kJ/mol
    3. C -80 kJ/mol
    4. D +130 kJ/mol
13. 131 mark

    An energy level diagram for a reaction shows the reactants at 210 kJ/mol, the peak of the curve at 330 kJ/mol, and the products at 150 kJ/mol.

    Which one of the following statements about this reaction is INCORRECT?

    1. A The reaction is exothermic, since the products (150 kJ/mol) are at a lower energy than the reactants (210 kJ/mol).
    2. B The activation energy of the forward reaction is 120 kJ/mol.
    3. C The overall enthalpy change for the reaction is -60 kJ/mol.
    4. D The activation energy of the reverse reaction is 120 kJ/mol.
14. 141 mark

    In a calorimetry experiment, a reaction transfers 1680 J of energy to 100 g of water. The specific heat capacity of water is 4.2 J per gram per degree C.

    Calculate the temperature rise of the water.

    1. A 400 degrees C
    2. B 4 degrees C
    3. C 16.8 degrees C
    4. D 0.25 degrees C
15. 151 mark

    In a reaction, the bonds broken in the reactants are 4 C-H bonds (413 kJ/mol each) and 2 O=O bonds (498 kJ/mol each). The bonds formed in the products are 2 C=O bonds (799 kJ/mol each) and 4 O-H bonds (463 kJ/mol each).

    Calculate the overall energy change for this reaction, and state whether it is exothermic or endothermic.

    1. A -802 kJ/mol; the reaction is exothermic.
    2. B +802 kJ/mol; the reaction is endothermic.
    3. C -351 kJ/mol; the reaction is exothermic.
    4. D -1798 kJ/mol; the reaction is exothermic.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: C

   1. Moles of sodium hydroxide = concentration x volume (in dm3) = 0.30 mol/dm3 x (20.0 / 1000) dm3 = 0.30 x 0.020 = 0.006 mol.
   2. Sodium hydroxide and hydrochloric acid react in a 1:1 mole ratio, so moles of hydrochloric acid = moles of sodium hydroxide = 0.006 mol.
   3. Concentration of the hydrochloric acid = moles / volume = 0.006 mol / (30.0 / 1000) dm3 = 0.006 / 0.030 = 0.2 mol/dm3.
   4. So the concentration of the hydrochloric acid is 0.2 mol/dm3, option C.
   • Why not A: Swaps the two volumes when scaling the known concentration, computing 0.30 x (30/20) = 0.45, the ratio of the acid's own volume to the base's volume, rather than the base's volume to the acid's volume.
   • Why not B: Adds the two volumes together into a single combined volume (20 + 30 = 50 cm3) and divides the moles of sodium hydroxide by that total, instead of dividing by the acid's own volume alone.
   • Why not D: Wrongly assumes hydrochloric acid is diprotic, like sulfuric acid, and so halves the correctly calculated concentration to account for a second acidic hydrogen that HCl does not actually have.

2. Question 2Answer: A

   1. A change of 1 on the pH scale corresponds to a change in H+ ion concentration by a factor of 10, so a fall in concentration to 1/1000 of its original value is a fall by a factor of 10 x 10 x 10.
   2. A factor of 10^3 in the H+ concentration corresponds to a change of 3 pH units.
   3. Diluting an acid lowers its H+ concentration, so its pH increases: new pH = original pH + 3 = 1 + 3.
   4. So the diluted solution has pH 4, option A.
   • Why not B: Mistakes the dilution factor itself (1000) for the new pH value, rather than converting that factor into a change in pH.
   • Why not C: Correctly converts the factor of 1000 into a pH change of 3 units, but stops there and gives that change itself as the new pH, forgetting to add it to the original pH of 1.
   • Why not D: Subtracts the pH change from the original pH instead of adding it, reversing the direction of the effect: diluting an acid makes it less concentrated in H+ ions, so its pH must rise, not fall.

3. Question 3Answer: D

   1. An amphoteric oxide is one that can react chemically as both a base and an acid, depending on what it is reacted with.
   2. Aluminium oxide reacts with acids to form an aluminium salt and water, behaving as a base: Al2O3 + 6HCl -> 2AlCl3 + 3H2O.
   3. Aluminium oxide also reacts with alkalis such as sodium hydroxide solution, behaving as an acid by being neutralised to form a soluble aluminate salt and water.
   4. So aluminium oxide reacts with both acids and alkalis, option D.
   • Why not A: Correctly identifies that it reacts with acids, as a base does, but wrongly claims it cannot also react with alkalis, missing the defining feature of an amphoteric oxide.
   • Why not B: Misreads 'amphoteric' as meaning unreactive, when the term in fact describes an oxide capable of reacting as both an acid and a base, not one that reacts as neither.
   • Why not C: Correctly identifies that it reacts with alkalis, as an acidic oxide would, but wrongly claims it cannot also react with acids, missing that it behaves both ways.

4. Question 4Answer: B

   1. A base is defined as a substance that can form OH-(aq) ions, or, equally, as a substance that can accept H+ ions; these are two routes to the same classification, and a substance need only satisfy one of them.
   2. Copper(II) oxide does not dissolve in water, so it cannot supply OH-(aq) ions that way, but it does react with an acid, for example CuO + 2HCl gives CuCl2 + H2O, accepting H+ ions from the acid as it reacts.
   3. That reaction alone is sufficient to classify copper(II) oxide as a base, even though it is insoluble and so is not an alkali.
   4. So the student is not correct: copper(II) oxide is a base by the H+-accepting definition, option B.
   • Why not A: Restricts the definition of a base to only the OH--forming route, when a base can equally be defined as a substance that accepts H+ ions, a route that does not require the substance to dissolve at all.
   • Why not C: Wrongly equates 'base' with 'alkali'; an alkali is specifically a base that is soluble in water and forms OH-(aq) ions, so calling every metal oxide an alkali overstates what insoluble oxides do.
   • Why not D: Wrongly restricts which metals can form basic oxides to only the most reactive groups, when a transition metal oxide such as copper(II) oxide is a perfectly ordinary base, reacting with acids to form a salt and water.

5. Question 5Answer: A

   1. Nitric acid reacting with potassium hydroxide is a neutralisation reaction, forming potassium nitrate and water.
   2. The measured rise in temperature shows energy has been released to the surroundings, so the reaction is exothermic.
   3. On an energy level diagram, an exothermic reaction is drawn with the products at a lower energy than the reactants, since the reaction as a whole has released energy overall.
   4. So the observation is consistent with the products being drawn lower than the reactants, option A.
   • Why not B: Reverses which side of the diagram is higher, and confuses a release of energy to the surroundings (which is what actually raises the surroundings' temperature) with an absorption of energy by the reacting mixture.
   • Why not C: Wrongly infers that a fast, easily observed temperature change means no energy barrier exists at all; a reaction can still have a genuine activation energy and proceed quickly once particles collide with enough energy to clear it.
   • Why not D: Wrongly invokes equilibrium to argue the reactants and products carry the same energy; a measured, sustained temperature change is direct evidence that the products and reactants do not carry the same energy here.

6. Question 6Answer: D

   1. Pressure affects the rate of a reaction only when a reactant is a gas, since pressure changes the concentration of a gas by compressing or expanding it.
   2. In this reaction neither reactant is a gas: magnesium is a solid and sulfuric acid is in aqueous solution, so changing the surrounding pressure does not change the concentration of either reactant.
   3. Temperature, acid concentration and surface area, by contrast, all genuinely increase the rate of this particular reaction, by the usual collision-theory mechanisms.
   4. So the change that would NOT increase the rate here is increasing the surrounding pressure, option D.
   • Why not A: Increasing temperature does increase the rate, by increasing both the frequency of collisions and the proportion with enough energy to react, so this is not a change that fails to increase the rate.
   • Why not B: Increasing the acid's concentration does increase the rate, by increasing the frequency of collisions between acid particles and the magnesium surface, so this is not the correct choice either.
   • Why not C: Cutting the ribbon into smaller pieces does increase the rate, by exposing a greater surface area of magnesium to the acid, so this too is a change that increases rate, not one that fails to.

7. Question 7Answer: A

   1. The rate of a reaction can be tracked by measuring the loss of a reactant, the gain of a product, or a physical property that changes as the reaction proceeds, whichever is practical for the reaction in question.
   2. Here the reaction produces a precipitate of sulfur that makes the mixture progressively cloudier, a physical property that is easy to observe without any special apparatus.
   3. Timing how long it takes for a cross viewed through the mixture to become obscured gives a direct, practical measure of how quickly the cloudiness develops.
   4. So the appropriate method here is timing the cross's disappearance as the mixture clouds over, option A.
   • Why not B: Ignores that the sulfur dioxide gas produced here is only a small, hazardous trace that is not practical or safe to collect and measure directly, unlike a reaction producing a substantial, harmless volume of gas.
   • Why not C: Confuses a genuine colour change, which a colorimeter is designed to detect, with the cloudiness (turbidity) actually produced here; the mixture does not turn a distinct colour, it becomes opaque with suspended sulfur particles.
   • Why not D: Sealing the flask traps the small amount of gas produced inside it, so almost no mass would be lost even if this method were otherwise appropriate, and in any case only a trace of gas escapes here.

8. Question 8Answer: C

   1. The mean rate of reaction over an interval is the change in the measured quantity divided by the length of that time interval.
   2. The volume changed from 10 cm3 at 10 seconds to 40 cm3 at 25 seconds, a change of 40 - 10 = 30 cm3.
   3. The time interval is 25 - 10 = 15 seconds.
   4. Mean rate = 30 / 15 = 2 cm3/s, option C.
   • Why not A: Divides the total volume collected by 25 seconds (40 cm3) by the total elapsed time since the very start of the experiment (25 s), rather than using only the volume produced and time elapsed during the stated interval.
   • Why not B: Inverts the calculation, dividing the time interval by the volume change (15 / 30) instead of dividing the volume change by the time interval.
   • Why not D: Divides the correct volume change of 30 cm3 by the total elapsed time of 25 s since the experiment began, rather than by the 15 s length of the interval over which that volume change actually occurred.

9. Question 9Answer: A

   1. Collision theory explains rate in terms of both how often particles collide and what proportion of those collisions have enough energy to react.
   2. Doubling the acid's concentration means there are twice as many acid particles in the same volume of solution, so they collide with the magnesium surface twice as often in a given time.
   3. The temperature is unchanged, so the average kinetic energy of the particles, and therefore the proportion of collisions with energy at or above the activation energy, is unaffected by this change.
   4. So the increase in rate here comes purely from a greater frequency of collisions, option A.
   • Why not B: Confuses the effect of concentration with the effect of temperature; it is raising the temperature, not the concentration, that increases the average kinetic energy of particles and so the proportion of collisions exceeding the activation energy.
   • Why not C: Confuses the effect of concentration with the effect of a catalyst; only a catalyst provides an alternative pathway that lowers the activation energy, concentration does not change it.
   • Why not D: Overlooks that the acid's own concentration is itself a rate factor here, since it changes how often acid particles collide with the magnesium surface, independent of the fixed mass of magnesium or the unchanged temperature.

10. Question 10Answer: D

    1. The activation energy is the difference between the energy of the reactants and the energy at the peak of the curve.
    2. Before the catalyst: activation energy = 250 - 80 = 170 kJ/mol.
    3. After the catalyst: activation energy = 190 - 80 = 110 kJ/mol.
    4. Decrease = 170 - 110 = 60 kJ/mol, option D.
    • Why not A: Gives the original, uncatalysed activation energy itself (250 - 80 = 170 kJ/mol), rather than the decrease caused by the catalyst.
    • Why not B: Gives the new, catalysed activation energy itself (190 - 80 = 110 kJ/mol), rather than the decrease between the two activation energies.
    • Why not C: Reads off the new peak height alone (190 kJ/mol), without subtracting the energy of the reactants at all, so this is not an activation energy or a decrease in one.

11. Question 11Answer: B

    1. A catalyst provides an alternative reaction pathway with a lower activation energy for both the forward and reverse reactions.
    2. Because both directions are sped up equally, a catalyst does not affect the position of equilibrium, and so does not affect the equilibrium yield of product.
    3. What a catalyst does change is how quickly that equilibrium is reached, since both the forward and reverse reactions now proceed faster than before.
    4. So the equilibrium is reached sooner, but the yield of product at equilibrium is the same either way, option B.
    • Why not A: Wrongly assumes a catalyst speeds up the forward and reverse reactions unequally; a catalyst lowers the activation energy of both directions by the same amount, so neither is favoured over the other.
    • Why not C: Wrongly assumes a catalyst favours the reverse reaction; by the same reasoning it cannot favour either direction, so it cannot shift the yield towards reactants either.
    • Why not D: Correctly implies no change to the yield, but wrongly denies the catalyst has any effect on rate at all; a catalyst does speed up how quickly equilibrium is reached, even though it leaves the final yield unchanged.

12. Question 12Answer: C

    1. Hess's Law states that the overall enthalpy change for a reaction is the same regardless of the route taken, provided the starting and finishing substances are the same.
    2. Here X can be converted into Z either directly (delta H1) or via Y in two steps (delta H2 then delta H3), so delta H1 = delta H2 + delta H3.
    3. delta H1 = (+130) + (-210) = -80 kJ/mol.
    4. So the direct conversion of X into Z has an enthalpy change of -80 kJ/mol, option C.
    • Why not A: Adds the magnitude of the second step without keeping its negative sign, treating the exothermic step of -210 kJ/mol as though it were +210 kJ/mol, giving 130 + 210 = 340 instead of 130 + (-210).
    • Why not B: Uses only the enthalpy change of the second step (Y -> Z) and ignores the first step (X -> Y) altogether, rather than adding the two steps together as Hess's Law requires.
    • Why not D: Uses only the enthalpy change of the first step (X -> Y) and ignores the second step (Y -> Z) altogether, rather than adding the two steps together as Hess's Law requires.

13. Question 13Answer: D

    1. The forward activation energy is the peak minus the reactants: 330 - 210 = 120 kJ/mol, so statement B is true.
    2. The overall enthalpy change is the products minus the reactants: 150 - 210 = -60 kJ/mol, a negative value, so the reaction is exothermic and statement A and statement C are both true.
    3. The reverse activation energy is the peak minus the products, not minus the reactants: 330 - 150 = 180 kJ/mol, not 120 kJ/mol as statement D claims.
    4. So the incorrect statement is D, which has used the reactants' energy instead of the products' energy when finding the reverse activation energy.
    • Why not A: This statement is actually true: products (150) lower than reactants (210) is precisely what defines an exothermic reaction, so choosing it as the incorrect statement mistakes a correct description for a false one.
    • Why not B: This statement is true: the forward activation energy is the peak minus the reactants, 330 - 210 = 120 kJ/mol, so choosing it as the incorrect statement means this correct subtraction was not carried out or checked.
    • Why not C: This statement is true: the overall enthalpy change is the products minus the reactants, 150 - 210 = -60 kJ/mol, so choosing it as the incorrect statement means this correct subtraction was not carried out or checked.

14. Question 14Answer: B

    1. Specific heat capacity is defined as energy transferred divided by (mass x temperature change), so temperature change = energy transferred / (mass x specific heat capacity).
    2. Here energy transferred = 1680 J, mass = 100 g, specific heat capacity = 4.2 J per gram per degree C.
    3. Mass x specific heat capacity = 100 x 4.2 = 420, so temperature change = 1680 / 420.
    4. 1680 / 420 = 4, so the temperature rise of the water is 4 degrees C, option B.
    • Why not A: Divides the energy by the specific heat capacity only (1680 / 4.2), omitting the mass of water entirely from the calculation.
    • Why not C: Divides the energy by the mass only (1680 / 100), omitting the specific heat capacity entirely from the calculation.
    • Why not D: Inverts the rearrangement, computing mass x specific heat capacity divided by energy (100 x 4.2 / 1680) instead of energy divided by mass x specific heat capacity.

15. Question 15Answer: A

    1. Bond breaking is endothermic and bond formation is exothermic, so the overall energy change equals the energy needed to break the reactants' bonds minus the energy released forming the products' bonds.
    2. Energy to break bonds = (4 x 413) + (2 x 498) = 1652 + 996 = 2648 kJ/mol.
    3. Energy released forming bonds = (2 x 799) + (4 x 463) = 1598 + 1852 = 3450 kJ/mol.
    4. Overall energy change = 2648 - 3450 = -802 kJ/mol, a negative value, so the reaction is exothermic overall, option A.
    • Why not B: Reverses the standard order of subtraction, computing energy released forming bonds minus energy absorbed breaking bonds (3450 - 2648 = +802) instead of the other way round, and then mislabels this reversed, positive value as endothermic.
    • Why not C: Adds the bond energies given without multiplying by the stated number of each type of bond present (413 + 498 = 911 broken; 799 + 463 = 1262 formed; 911 - 1262 = -351), understating both totals by ignoring how many of each bond actually break or form.
    • Why not D: Omits the 2 O=O bonds broken from the total energy required to break bonds, using only the 4 C-H bonds (4 x 413 = 1652) while still using the full, correct total energy released forming bonds (3450), giving 1652 - 3450 = -1798.