At temperatures above 5°C, the benzenediazonium ion, C₆H₅N₂⁺, reacts with water as shown. C₆H₅N₂⁺(aq) + H₂O(l) → C₆H₅OH(aq) + N₂(g) + H⁺(aq) A student investigates this reaction by measuring the volume of nitrogen gas produced at regular time intervals. The diagram shows the experimental set-up used to investigate this reaction. [Figure showing a conical flask with C₆H₅N₂⁺(aq) connected to a gas syringe]
- (a)The student finds that the reaction is very slow, so decides to investigate the reaction at 30°C. Complete the diagram to show how the student could investigate this reaction at a constant 30°C.[2]
- (b)(i)The student prepared a solution of C₆H₅N₂⁺(aq) at 5°C. A 200.0 cm³ sample of this solution was placed in a conical flask. The apparatus was allowed to equilibrate at 30°C. The gas syringe was then connected, a stop-clock was started and readings of time and gas volume were taken. When the decomposition of the C₆H₅N₂⁺ ion was complete (as shown by no more gas production) the final volume of N₂(g) produced, Vfinal, was 72 cm³. Show by calculation that when the stop-clock was started the concentration of C₆H₅N₂⁺(aq) was 0.0150 mol dm⁻³. [The molar volume of gas under room conditions is 24.0 dm³.] concentration of C₆H₅N₂⁺(aq) = ........................................... mol dm⁻³[2]
- (b)(ii)The volume of nitrogen produced is proportional to the amount of C₆H₅N₂⁺ that reacts. As C₆H₅N₂⁺ reacts, its concentration in the solution falls. Using the axes shown, sketch a graph to show the change in volume of N₂(g) produced with the change in concentration of C₆H₅N₂⁺(aq). Label the axes [C₆H₅N₂⁺(aq)]/mol dm⁻³ and volume of N₂(g)/cm³, putting the independent variable on the x-axis. Include on the axes the maximum values for concentration and volume where the lines on the axes are shown. [Figure showing blank axes for plotting][2]
- (c)The results the student obtained are shown in the table. To calculate the concentration of C₆H₅N₂⁺(aq) the student used the following equation. [C₆H₅N₂⁺(aq)] = 0.0150 × (1 – V/Vfinal) V = volume of N₂(g) recorded at a specified time. Vfinal = final volume of N₂(g) at complete decomposition of C₆H₅N₂⁺. Vfinal is 72 cm³. Complete columns C and D to three significant figures. You may use the space below for any working. [Table with columns A (time/minutes), B (volume of N₂(g), V/cm³), C (V/Vfinal), D ([C₆H₅N₂⁺(aq)]/mol dm⁻³)][2]
- (d)Plot a graph on the grid to show the relationship between [C₆H₅N₂⁺(aq)] and time. Use a cross (×) to plot each data point. Draw a line of best fit. [Grid for plotting provided][2]
- (e)Draw a tangent at time = 0.0 on your graph and use the tangent to determine the initial rate of reaction. State the units of the initial rate of reaction. State the co-ordinates of both points you used in your calculation. co-ordinates 1 .................................................. co-ordinates 2 .................................................. initial rate of reaction = ................................................... units = ...................................................[4]
- (f)Use your graph to complete the table and calculate the values of two half-lives. For the first half-life, concentrations have been chosen for you. [Table with columns concentration 1, time 1, concentration 2, time 2, t1/2][3]
- (g)State and explain what the values of the two half-lives suggest about the order of reaction with respect to [C₆H₅N₂⁺(aq)].[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [1] Water bath/beaker of water containing thermometer around flask
- [1] Controlled heat source or heater/temperature regulator
- [1] Moles N2 = 72 / 24 000 = 0.003 moles
- (1 mol C6H5N2+Cl¯ → 1 mol N2)
- [1] Moles C6H5N2+ in 1000 cm³ solution = 0.003 × (1000 / 200) = 1.50 × 10¯² (mol)
- [1] Axes (label with quantity or correct unit) and values correct
- [1] Straight line from axis marks OR from 0,0 over most of the axes
- [1] Column values for D correctly calculated
- [1] 3 sf in C and D
- [1] Candidate's calculated points correctly plotted from table in 2(c)
- [1] Smooth curve of best fit
- [1] Tangent drawn at time zero
- [1] 2 sets of co-ordinates shown
- [1] calculation of gradient of tangent
- [1] mol dm⁻³ minute(s)⁻¹
- [1] Columns 1 and 3
- [1] Columns 2 and 4
- [1] Half-lives correctly calculated.
- [1] First order AND because half-lives are constant/equal
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