Benzenediazonium chloride, C6H5N2Cl, decomposes in water as shown in the following equation. C6H5N2Cl(aq) + H2O(l) → C6H5OH(aq) + N2(g) + HCl(aq) A solution of 0.0750moldm¬³ of C6H5N2Cl(aq) decomposes at a constant temperature of 50°C. The volume of nitrogen gas, N2(g), collected is recorded every 5 minutes for 45 minutes.
- (a)Draw a labelled diagram to show how the apparatus could be set up to carry out this experiment.[3]
- (b)(i)On Fig. 2.1, draw a tangent to the curve at t = 0 mins. Calculate the initial rate of reaction in cm³ min-1. initial rate of reaction = .............................................................. cm³ min-1[2]
- (b)(ii)Explain why the initial rate of reaction is calculated at t = 0 mins rather than dividing the total volume of gas produced by the time taken to produce it.[1]
- (b)(iii)Describe how the curve in Fig. 2.1 would be different, if at all, if the atmospheric pressure increases. All other conditions stay the same.[1]
- (b)(iv)On Fig. 2.1, draw a second curve to show the graph produced if the same volume of 0.0375 moldm-3 C6H5N2Cl(aq) decomposes at a constant temperature of 50°C. All other conditions stay the same.[1]
- (c)(i)Complete the table to calculate the values for the rate of N₂ production. Give your answers to three significant figures. Table 2.1 concentration of C6H5N2Cl(aq) / moldm-3 time taken to collect 0.0150 dm³ of N₂/s rate of N2 production /dm³s-1 0.500 21 0.400 33 0.300 48 0.200 64 0.100 122 [1]
- (c)(ii)The reaction is first order and obeys the following rate equation. rate = k [C6H5N2Cl] Explain how the data in Table 2.1 supports this statement.[1]
- (c)(iii)State the dependent variable in this investigation.[1]
- (c)(iv)The student wants to perform a similar experiment using 0.200 moldm¯³ C6H5N2Cl(aq). Describe how the student should make a standard solution of 100.0 cm³ of 0.200 mol dm-3 C6H5N2Cl(aq) starting from a solution of 0.500 moldm−3 C6H5N2Cl(aq). Give the name and size of any key apparatus which should be used and describe how the student should ensure the volume is exactly 100.0cm³. Write your answer using a series of numbered steps.[3]
- (c)(v)Explain why (iv) must be carried out at a temperature below 5°C.[1]
- (d)(i)The student's final yield was 52%. They had not spilled any reagents or products. Suggest two reasons why the student's yield was lower than 100%. Assume no errors were made in the measurement of any substances. reason 1: reason 2:[2]
- (d)(ii)Explain why the solid dye should be dried before assessing the melting point.[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 (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [3] M1: Leakproof reaction vessel capable of delivering a sample of gas. M2: Suitable method of collection and measurement of the gas produced. (eg. over water or gas syringe). M3: method for monitoring & maintaining temperature (thermostatically controlled water bath).
- [2] M1: A tangent line which is... • a straight line • starts at / passes through 0,0. • keeps to the left of the curve of the original line. • covers at least 25 cm³ (5 big squares) on y-axis. • passes between 4.0 and 5.5 minutes (inc.) at 25 cm³ M2: correct calculation of the gradient of tangent line drawn.
- [1] The total volume of gas produced divided by the time taken is the mean rate (rather than the initial rate).
- [1] It (the line / curve) would be below / lower (than the one on the graph) (starting at 0,0).
- [1] A second curve which: • starts at 0,0 (± 1 small square) • is less steep than curve for 0.0750 moldm⁻³. (before any plateau) • produces 27.5 cm³ gas in total by 45 minutes. (± half a small square). • Extends from 0 – 45 minutes.
- [1] concentration rate 0.5 0.000714 7.14 ×10⁻⁴ 0.4 0.000455 4.55 ×10⁻⁴ 0.3 0.000313 3.13 ×10⁻⁴ 0.2 0.000234 2.34 ×10⁻⁴ 0.1 0.000123 1.23 ×10⁻⁴
- [1] explanation using numbers from the table that whatever change is made to the concentration causes the same change in the rate (within experimental error).
- [1] time (taken to collect 0.0150dm³ of N2)
- [3] M1: Correct volume, 40(.0) cm³, (of 0.5(00) moldm⁻³ ) C6H5N2Cl(aq) is used / measured. M2: measure the volume from M1 C6H5N2CI using a (50 cm³) burette into a 100 cm³ volumetric flask. M3: make up to the (calibration) mark with distilled water. AND then mix the solution (by inverting the flask)
- [1] to make sure the concentration does not change (from 0.200 moldm³) (by making sure the C6H5N2CI does not decompose)
- [2] Two acceptable reasons for low yield. 1 Some C6H5N2CI had decomposed / reacted with water (before reacting with phenylamine) OR temperature rose above 5°C (in step 1) OR phenylamine used was impure. 2 Some dye / solid is not recovered / is lost through / left in filter OR some of the dye dissolves in the water / is aqueous (in step 2).
- [1] (Residual) water causes solid to have a different / inaccurate (lower) melting point (to the dry solid). OR To ensure all water has been driven off to increase accuracy in measurement of melting point
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