May/June 2026 Paper 52

2026 · 2 questions · 21 parts · 30 marks

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1(a)(i)1(a)(ii)1(b)(i)1(b)(ii)1(b)(iii)1(b)(iv)1(c)(i)1(c)(ii)1(c)(iii)1(c)(iv)1(c)(v)

–/16 mks

2(a)2(b)2(c)2(d)2(e)(i)2(e)(ii)2(e)(iii)2(e)(iv)2(e)(v)2(e)(vi)

–/14 mks

Context for question 2

This question is about the rate of the reaction between aqueous sodium ethanedioate,

N a_{2} C_{2} O_{4}

(aq), and aqueous sodium peroxydisulfate,

N a_{2} S_{2} O_{8}

(aq).

N a_{2} C_{2} O_{4} (a q) + N a_{2} S_{2} O_{8} (a q) \to 2 N a_{2} S O_{4} (a q) + 2 C O_{2} (g)

When

N a_{2} C_{2} O_{4}

(aq) is in excess, the rate equation for this reaction is:

r a t e = k [N a_{2} S_{2} O_{8}]^{m}

where

k

is the rate constant and

m

is the order of the reaction with respect to

N a_{2} S_{2} O_{8}

(aq).

In an experiment, excess

N a_{2} C_{2} O_{4}

(aq) is added to a sample of

N a_{2} S_{2} O_{8}

(aq) in a side-arm conical flask. The volume of carbon dioxide gas,

C O_{2}

(g), produced is measured at regular time intervals.

Figure 2.1 shows the results from the experiment plotted on the grid. A curved line of best fit has been drawn.

Figure 2.1, a graph of volume of $CO_2$(g) produced in cm$^{3}$ (y-axis, 0 to 12) against time in s (x-axis, 0 to 1600), with eight plotted crosses joined by a single smooth curve of best fit that rises steeply from the origin, passing close to (200, 3.8), (400, 6.1), (600, 7.5) and (800, 8.4), then flattening through (1200, 9.6) and (1400, 10.0) to level out at 10 cm$^{3}$ by 1600 s; the plotted point near 1000 s sits noticeably below the curve, the anomalous point

2(a)Practical Techniques SafetyMedium2 marks

Complete Figure 2.2 to show how the apparatus should be set up for measurement of the volume of

C O_{2}

(g) produced.

C O_{2}

(g) is soluble in water. Label your diagram.

Draw with the mouse or a finger.

2(b)Experimental PlanningEasy1 mark

Identify the independent variable.

2(c)Uncertainty Errors EvaluationEasy1 mark

Suggest a timing error that could have led to the anomalous point in Figure 2.1.

2(d)Rates ReactionMedium1 mark

On Figure 2.1, draw a second curve to show the graph produced if the experiment is done at a higher temperature. All other conditions stay the same.

Draw with the mouse or a finger.

Context for 2(e)

A student carries out the same procedure using different concentrations of

N a_{2} S_{2} O_{8}

. The results are used to produce the data shown in Table 2.1.

2(e)(i)Tables Graphs Data AnalysisEasy2 marks

Complete Table 2.1. Record your values to two decimal places.

[ $N a_{2} S_{2} O_{8}$ ] / mol dm $^{- 3}$	rate / mol dm $^{- 3}$ s $^{- 1}$	$lo g [N a_{2} S_{2} O_{8}]$	log(rate)
$1.00 \times 1 0^{- 2}$	0.02760	Table answer r1_$\log[Na_2S_2O_8]$	Table answer r1_log(rate)
$5.74 \times 1 0^{- 3}$	0.01160	Table answer r2_$\log[Na_2S_2O_8]$	Table answer r2_log(rate)
$3.67 \times 1 0^{- 3}$	0.00549	Table answer r3_$\log[Na_2S_2O_8]$	Table answer r3_log(rate)
$2.88 \times 1 0^{- 3}$	0.00398	Table answer r4_$\log[Na_2S_2O_8]$	Table answer r4_log(rate)
$2.08 \times 1 0^{- 3}$	0.00250	Table answer r5_$\log[Na_2S_2O_8]$	Table answer r5_log(rate)
$1.57 \times 1 0^{- 3}$	0.00169	Table answer r6_$\log[Na_2S_2O_8]$	Table answer r6_log(rate)

2(e)(ii)Tables Graphs Data AnalysisMedium2 marks

Use the results from Table 2.1 to plot a graph on the grid in Figure 2.3 to show the relationship between

lo g [N a_{2} S_{2} O_{8}]

and log(rate). Use a cross (

\times

) to plot each data point. Draw a straight line of best fit.

Draw with the mouse or a finger.

2(e)(iii)Tables Graphs Data AnalysisMedium2 marks

Determine the gradient of your line of best fit in Figure 2.3. State the coordinates of both points you use in your calculation. These must be selected from your line of best fit. Give the gradient to three significant figures.

coordinates 1 and coordinates 2 (from your line of best fit)

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gradient =

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2(e)(iv)Rate Equations Reaction MechanismsMedium1 mark

The rate equation can also be written as shown.

lo g (r a t e) = m lo g [N a_{2} S_{2} O_{8}] + lo g k

Use your answer to 2(e)(iii) to deduce a value for

k

. [If you were unable to find the gradient in 2(e)(iii), then use the value 1.55. This may not be the correct answer.]

k =

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2(e)(v)Uncertainty Errors EvaluationHard1 mark

The true value for the order of reaction,

m

, is

\frac{3}{2}

. The total percentage error from measurements is determined to be 4.5%. Use this and your gradient from 2(e)(iii) to determine whether the error in the experiment could be accounted for by error from measurements or is caused by other factors. [If you were unable to find the gradient in 2(e)(iii), then use the value 1.55. This may not be the correct answer.] Show your working.

Answer

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2(e)(vi)Uncertainty Errors EvaluationEasy1 mark

Use your graph in Figure 2.3 to state whether the results from the experiment are reliable. Give a reason for your answer.

Answer

0 words