Rates of Reaction & Energetics

11 methods · 24 questions

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On a 'which line' question that plots volume or mass of product against time, read the curve as two independent things: its final plateau and its initial steepness. They are set by different quantities, so decide each one separately.

Trigger: a graph comparing experiments, or an experiment described against a reference line P, on a product-versus-time plot.

Instances:
(i) the plateau height is fixed only by the moles of the limiting reagent: work out the moles of each reactant against the equation ratio, the smaller caps the product;
(ii) the initial steepness is fixed by the rate factors: concentration, surface area (powder faster than granules faster than a lump), temperature and catalyst;
(iii) a change that only affects rate (surface area, temperature, catalyst, mass of catalyst) leaves the plateau exactly where it was, so both curves finish at the same height;
(iv) for a 'produce this amount in the shortest time' question, first eliminate every option that cannot make that amount at all, then compare the rate of the survivors.

Linked questions (3)

NSAA 2016 Section 1, Part C (Chemistry questions) · Q44

The graphs show results of two experiments (W and Z) involving the catalytic decomposition of hydrogen peroxide.

Assuming all other conditions are kept constant, which one of the following options would lead to the results shown?

The options list, in order: the conditions of experiment W and the conditions of experiment Z.

A graph of total volume of oxygen (vertical axis) against time (horizontal axis), with arrowheads on both axes and no numerical scales. Two curves start at the origin. The curve labelled experiment Z rises steeply and levels off quickly at a lower final volume. The curve labelled experiment W rises less steeply, keeps rising for longer and levels off at a final volume roughly twice that of experiment Z. Each plateau is extended by a thin horizontal line to its label on the right.

A $100 cm^{3}$ of $1.0 mol dm^{- 3}$ hydrogen peroxide; $50 cm^{3}$ of $2.0 mol dm^{- 3}$ hydrogen peroxide
Bcatalyst is in lumps; catalyst is finely divided
Creaction carried out at $25^{\circ} C$ ; reaction carried out at $50^{\circ} C$
D $2.0 g$ manganese(IV) oxide used; $1.0 g$ manganese(IV) oxide used
E $100 cm^{3}$ of $1.0 mol dm^{- 3}$ hydrogen peroxide; $25 cm^{3}$ of $2.0 mol dm^{- 3}$ hydrogen peroxide

Worked solution

Answer:E

Two features of the graphs must be explained. Experiment Z has the steeper initial gradient, so its initial rate is higher. Experiment Z also levels off at a smaller total volume of oxygen, about half that of experiment W, so it must contain fewer moles of hydrogen peroxide.

Options B, C and D change only the rate (surface area of the catalyst, temperature, mass of catalyst); the amount of hydrogen peroxide would be the same in both experiments, so both curves would reach the same final volume. They are ruled out.

Option A: W contains

0.100 \times 1.0 = 0.10 mol

and Z contains

0.050 \times 2.0 = 0.10 mol

of hydrogen peroxide, so the final volumes would again be equal. Ruled out.

Option E: W contains

0.10 mol

and Z contains

0.025 \times 2.0 = 0.05 mol

, so Z gives half the volume of oxygen; and Z uses the more concentrated solution, so it reacts faster. Both features of the graphs are reproduced.

The answer is E.

NSAA 2017 Section 1, Part C (Chemistry questions) · Q48

Magnesium reacts with sulfuric acid according to the following chemical equation:

Mg (s) + H_{2} S O_{4} (aq) \to MgS O_{4} (aq) + H_{2} (g)

Line P on each graph shows how the volume of hydrogen formed changes with time when

1.2 g

of magnesium reacts with

40 cm^{3}

1.0 mol dm^{- 3}

sulfuric acid at

20^{\circ} C

.

(

A_{r}

value: Mg = 24)

Two further experiments were carried out and the volumes of hydrogen formed were plotted.

Experiment Q:

1.2 g

of magnesium +

40 cm^{3}

2.0 mol dm^{- 3}

sulfuric acid at

20^{\circ} C

Experiment R:

1.2 g

of magnesium +

40 cm^{3}

0.5 mol dm^{- 3}

sulfuric acid at

20^{\circ} C

Which lines show how the volume of hydrogen formed will change with time in each experiment?

The options list, in order: the line for experiment Q and the line for experiment R.

Two graphs side by side on square grids, each with volume of hydrogen / cm^3 on the vertical axis and time / s on the horizontal axis, both axes starting at 0. The left graph is titled experiment Q and shows four curves rising from the origin and levelling off: line P levels off at a certain height, line 1 rises more steeply and levels off at the same height as P, line 2 rises more steeply and levels off about a quarter higher than P, and line 3 rises most steeply and levels off at about twice the height of P. The right graph is titled experiment R and shows three curves rising from the origin: line P and line 4 level off at the same height, with line 4 rising less steeply than P, and line 5 rises least steeply and levels off at about half the height of P.

A1 ; 4
B1 ; 5
C2 ; 4
D2 ; 5
E3 ; 4
F3 ; 5

Worked solution

Answer:D

First work out which reactant limits the amount of hydrogen in each experiment. The magnesium is always

1.2/24 = 0.050 mol

, and the equation is 1 : 1.

Line P: acid

= 0.040 dm^{3} \times 1.0 = 0.040 mol

, less than the magnesium, so the acid is limiting and

0.040 mol

of hydrogen is made.

Experiment Q: acid

= 0.040 \times 2.0 = 0.080 mol

, now more than the magnesium, so the magnesium is limiting and

0.050 mol

of hydrogen is made,

1.25

times the volume of P. The more concentrated acid also makes the reaction faster, so the curve must rise more steeply than P and level off 25% higher. On the left graph, line 1 levels off at the same volume as P, line 3 at about twice the volume, and line 2 at about

1.25

times the volume: Q is line 2.

Experiment R: acid

= 0.040 \times 0.5 = 0.020 mol

, limiting, so only

0.020 mol

of hydrogen is made, half the volume of P, and the reaction is slower. On the right graph line 4 levels off at the same volume as P, while line 5 rises more slowly and levels off at half the volume: R is line 5.

The answer is D.

NSAA 2017 Section 1, Part C (Chemistry questions) · Q53

The reaction between calcium carbonate and hydrochloric acid was used to measure the effect of changing conditions on the mass of

C O_{2}

produced and the rate of

C O_{2}

production.

CaC O_{3} (s) + 2 HCl (aq) \to CaC l_{2} (aq) + H_{2} O (l) + C O_{2} (g)

The experiment was carried out five times with different conditions at a constant temperature.

The following conditions were varied:

CaC O_{3}

as chips or powder

mass of

CaC O_{3}

volume of HCl

concentration of HCl

Which experiment (A-E) in the following table will produce

8.8 g

of carbon dioxide in the shortest time?

(

M_{r}

values:

CaC O_{3} = 100

;

C O_{2} = 44

)

The options list, in order: the type of

CaC O_{3}

, the mass of

CaC O_{3}

in g, the volume of HCl in

cm^{3}

and the concentration of HCl in

mol dm^{- 3}

Achips ; 10 ; 400 ; 2.0
Bpowder ; 20 ; 100 ; 2.0
Cchips ; 20 ; 200 ; 2.0
Dpowder ; 10 ; 200 ; 2.0
Echips ; 20 ; 400 ; 1.0

Worked solution

Answer:C

First decide which experiments can produce

8.8 g

of carbon dioxide at all, then compare the rates of those that can.

8.8 g

C O_{2}

8.8/44 = 0.20 mol

. From the equation this needs

0.20 mol

CaC O_{3}

, that is

20 g

, and

0.40 mol

of HCl.

Experiments A and D use only

10 g

(

0.10 mol

) of calcium carbonate, so they can give at most

4.4 g

C O_{2}

. Experiment B has

20 g

of carbonate but only

0.100 dm^{3} \times 2.0 = 0.20 mol

of HCl, which is again enough for only

0.10 mol

C O_{2}

. These three are ruled out.

Experiment C has

20 g

of carbonate and

0.200 \times 2.0 = 0.40 mol

of HCl: exactly enough for

8.8 g

. Experiment E has

20 g

of carbonate and

0.400 \times 1.0 = 0.40 mol

of HCl: also exactly enough.

C and E both use chips of the same mass, so the only difference in rate comes from the acid concentration. The

2.0 mol dm^{- 3}

acid in C gives more frequent collisions than the

1.0 mol dm^{- 3}

acid in E, so C reaches

8.8 g

in the shorter time.

The answer is C.