A-LevelChemistryReaction kinetics9701/41 · Oct/Nov 2017 · Q112 Marks

The compound nitrosyl bromide, NOBr, can be formed by the reaction shown. 2NO + Br₂ ⇌ 2NOBr

(a)Using oxidation numbers, explain why this reaction is a redox reaction.[2]
(b)Nitrosyl bromide contains a trivalent nitrogen atom. Draw the 'dot-and-cross' diagram for NOBr. Show outer electrons only.[2]
(c)(i)What is meant by the term order of reaction with respect to a particular reagent?[1]
(c)(ii)Use the data in the table to deduce the values of a and b in the rate equation. Show your reasoning.[2]
(c)(iii)Use the data in the table to calculate the initial rate for experiment 4. initial rate = .............................. mol dm⁻³s⁻¹[1]
(c)(iv)Use the results of experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. rate constant, k = .............................. units ..............................[2]
(c)(v)By considering the rate equation, explain why the rate decreases with decreasing temperature.[1]
(d)The reaction between X and Y was studied. 2X + Y → Z The following sequence of steps is a proposed mechanism for the reaction. step 1 2X → V step 2 V + Y → Z The general form of the rate equation for this reaction is as follows. rate = k[X]ᵐ[Y]ⁿ Step 1 is the slower step in the mechanism. Deduce the values of m and n in the rate equation. m = .............................. n = ..............................[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

[1] (a) N +2 to +3 (and oxidised)
[1] (a) Br2/Br 0 to -1 (and reduced)
[1] (b) 3 bonding pairs around N (in a structure involving NOBr)
[1] (b) rest of molecule correct
[1] (c)(i) the power to which a concentration of a reactant is raised in the rate equation
[1] (c)(ii) using expt. 2 and 3
[1] (c)(ii) a = 2 or [NO] 2nd order and conc × 3 rate × 9 or 6.1 × 10⁻²/6.8 × 10⁻³ = (0.09/0.03)ᵃ
[1] (c)(ii) using expt. 1 and 2
[1] (c)(ii) b = 1 or [Br2] 1st order and conc × 2 rate × 2 or 6.8 × 10⁻³/3.4 × 10⁻³ = (0.04/0.02)ᵇ
[1] (c)(iii) initial rate = 0.16(32)
[1] (c)(iv) (0.0034 = k(0.03)²(0.02))
[1] (c)(iv) k = 188.9
[1] (c)(iv) mol⁻² dm⁶ s⁻¹
[1] (c)(v) k decreases (as rate decreases)
[1] (d) m = 2 and n = 0

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