Edexcel IAL Chemistry Topic 11 Kinetics

An answer that makes reference to the following point:

- rate $=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[I^{-}\right]$

Allow rate $=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[\left[^{-}\right]\left[\mathrm{H}^{+}\right]^{\circ}\right.$

Allow species in any order TE from (b)(i)

rearrangement of the rate equation and calculation of the rate constant

Example of calculation

$k=\operatorname{rate} \div\left(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right] \times\left[\mathrm{I}^{-}\right]\right)$$k=4.359 \times 10^{-6} \div\left(5.4 \times 10^{-2} \times 8.2 \times 10^{-3}\right)$$=4.359 \times 10^{-6} \div 4.428 \times 10^{-4}$$=9.844 \times 10^{-3}$

Ignore SF except 1SF

axes correct with scales allowing plot to cover at least half the grid and $\ln$ rate values becoming more negative down the axis

(1)

- both axes labelled with units $\mathrm{K}^{-1}$ on x axis

(1)

- points correctly plotted and best fit line drawn

(1)

- gradient in the range -7200 to -6800

(1)

- $\mathrm{E}_{\mathrm{a}}=-(-7000 \times 8.31) \div 1000$$$=+58.2\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$$

range 56.5 to 59.9 TE on incorrect gradient value but Ea

(1)

Example of calculation

Allow horizontal axis at bottom of graph Ignore extrapolation

Allow plotting of all points except the incorrect one

TE on incorrect gradient

Ignore SF

- correct value of 1/T from the graph: 0.00331 ± 0.00001 (1)
- conversion to temperature: 302 K; range 301–303 K (1)

TE from incorrect 1/T read from the graph. Ignore significant figures.

An answer that makes reference to the following points:

- calculation of one half-life (1)
- calculation of a second half-life which is constant / has a similar value, so the reaction is first order (1)

Example of calculation:
- t1 = 0.70 − 0 = 0.70 h
- t2 = 1.40 − 0.70 = 0.70 h, or t3 = 2.1 − 1.40 = 0.70 h

Additional guidance:
- Allow half-lives in the range 0.65–0.75 h
- Allow correct conversion into minutes, range 39–45 min
- Two correct half-lives without working and reference to constant value scores 1
- Some correct working must be shown on the graph; small dashes on the x-axis are sufficient

Either:

- no, because the order determined is with respect to ethyl butanoate only / water is a reactant (1)
- because water is in excess (1)

Or:

- yes, because water is in very large excess (1)
- so the concentration of water has no effect on the rate (1)

Additional guidance:
- Allow reactants need to be equimolar to determine overall order
- Allow that an additional experiment with ester in excess is needed to find the order with respect to water

An answer that makes reference to the following point:

- the reactants / NO and $\mathrm{O}_{2}$ are colourless but the product / $\mathrm{NO}_{2}$ is reddish brown / coloured

Allow just $\mathrm{NO} / \mathrm{O}_{2}$ is colourless and $\mathrm{NO}_{2}$ is brown

Allow just nitrogen dioxide / product is reddish brown / coloured / dark colour

Allow any combination of yellow, red, orange and brown for the colour of $\mathrm{NO}_{2}$

Allow measure the time for the brown gas to form

Ignore just 'there will be a colour change' / mixture will darken Ignore $\mathrm{NO}_{2}$ is a different colour form NO and $\mathrm{O}_{2}$

Do not award NO is coloured so there is a colour change

Do not award NO is yellow / red / orange / brown

(1)

rearrangement of rate equation expression and inserting values

- calculation of k and units

(1)

(1)

Example of calculation
$$=6.87 \times 10^{-4} \div\left(\left(6.50 \times 10^{-2}\right)^{2} \times 1.25 \times 10^{-2}\right)$$$=13.008 / 13.0 \mathrm{dm}^{6} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$

Correct answer with no working scores (2)

Correct numerical answer with incorrect units scores (1)

Allow units in any order

Allow $\mathrm{dm}^{6} / \mathrm{mol}^{2} \mathrm{~s}$$0.84554 / 0.846 \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$ (not squaring $6.50 \times 10^{-2}$ ) scores ( 1 ) for final value and units for M2

Ignore SF except 1SF

(2)

An answer that makes reference to the following points:

- adding the two steps together gives the overall equation

- the steps do not match the rate equation because the slow step should be the second step

Allow the two steps match the overall equation as the reactants and products are the same Allow $\mathrm{N}_{2} \mathrm{O}_{2}$ is formed then reacts / cancels out / is an intermediate Ignore just the overall equation is $2 \mathrm{NO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2}$

Allow it does not match because there is no oxygen in the slow step / rate determining step / rds Allow because in this mechanism oxygen is zero order / is not first order

Allow because with these steps the rate equation would be rate $=k[\mathrm{NO}]^{2}$

(2)