Nitrogen monoxide, NO(g), reacts with hydrogen, H2(g), under certain conditions. 2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
- (a)Define the term rate of reaction.[1]
- (b)Identify a change in the reaction mixture that would enable the rate of this reaction to be studied.[1]
- (c)The rate equation for this reaction is given. rate = k[NO]2[H2] The result of an experiment in which NO reacted with H₂ is shown in the table. initial [NO]/mol dm-3 | initial [H2]/moldm-³ | initial rate of reaction/mol dm-³ s-1 --- | --- | --- 2.50 x 10-3 | 2.50 x 10-3 | 1.27 x 10-3 Use the data and the rate equation to calculate a value for the rate constant k. Give the units of k.[2]
- (d)A second experiment is performed at the same temperature. The initial concentration of H2(g) is 4.60 x 10-3 mol dm-3. The initial rate of the reaction is 2.31 x 10-3 mol dm-3 s-1. Calculate the initial concentration of NO(g).[1]
- (e)State the order of the reaction with respect to NO(g) and with respect to H2(g), and the overall order of the reaction.[1]
- (f)(i)The reaction is believed to proceed in three steps. 1 2NO → N2O2 2 N2O2 + H₂ → N2O + H₂O 3 N2O + H2 → N₂ + H₂O Deduce which of the three steps is the rate-determining step.[1]
- (f)(ii)Explain your answer to (i).[1]
- (g)(i)A third experiment is performed under different conditions. A small amount of H2(g) of concentration 0.0200 moldm-³ is mixed with a large excess of NO(g). The concentration of H2(g) is found to have a constant half-life of 2.00 seconds under the conditions used. Define the term half-life.[1]
- (g)(ii)Use the axes below to construct a graph of the variation in the concentration of H2(g) during the first 6 seconds under the conditions used. [Figure showing blank graph axes for [H2] vs time][2]
- (h)(i)NO(g) acts as a catalyst in the oxidation of atmospheric sulfur dioxide. Give two equations to describe how NO(g) acts as a catalyst in this process.[1]
- (h)(ii)Explain why NO(g) can be described as a catalyst in this reaction.[1]
- (h)(iii)Describe, with the aid of an equation, an environmental consequence of the oxidation of atmospheric sulfur dioxide.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [1] change in amount / mass / concentration of reactant / product per time
- [1] decrease in volume or pressure
- [1] 8.13 x 104 / 81280 / 81300
- [1] mol−2 dm6 s−1
- [1] √(0.00231 / (0.0046 × 81280)) = 2.49 × 10−3
- [1] 2, 1, 3
- [1] 2
- [1] the total of steps 1 and 2 / the components of 2 are two NO and one H2
- [1] time for amount or mass or concentration to halve
- [1] 0.02 at start and 0.01 after 2 seconds
- [1] 0.005 after 4 seconds and 0.0025 after 6 seconds
- [1] NO + ½ O2 → NO2 or NO + O2 → NO2 + ½ O2 AND NO2 + SO2 → NO + SO3
- [1] (NO is) regenerated / reformed
- [1] SO3 + H2O → H2SO4 AND acid rain or consequence of this described
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →