[Maximum number: 21]

Phenacyl chloride has been used as a component of some tear gases. Its lachrymatory and irritant properties are due to it reacting with water inside body tissues to produce hydrochloric acid.

It undergoes a nucleophilic substitution reaction with NaOH(aq).

(a)

(i)

What is meant by the term order of reaction?

[ 7 ]

(ii)

Use the above data to deduce the order with respect to each reactant. Explain your reasoning.

[ 7 ]

(iii)

Write the overall rate equation for the reaction.

[ 7 ]

(a)

NO reacts readily with oxygen.

2 \mathrm{NO}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})

The table shows how the initial rate of this reaction at $25^{\circ} \mathrm{C}$ depends on the initial concentrations of the reactants.

[ 5 ]

(i)

Deduce the order of reaction with respect to each reactant. Explain your reasoning. order with respect to $[\mathrm{NO}(\mathrm{g})]$
order with respect to $\left[\mathrm{O}_{2}(\mathrm{~g})\right]$

[ 2 ]

(ii)

State the rate equation for this reaction. Use the rate equation to calculate the rate constant. Include the units for the rate constant in your answer.
rate =
rate constant, k=
units of k=

[ 3 ]

[Maximum number: 8]

The rate of the reaction $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is studied.

(a)

A small amount of $\mathrm{H}_{2}(\mathrm{~g})$ is mixed with a large excess of $\mathrm{I}_{2}(\mathrm{~g})$ at a temperature of 400 K and the reaction is monitored. The graph obtained is shown.

[ 2 ]

(i)

The reaction is first order with respect to $\mathrm{H}_{2}(\mathrm{~g})$ .

Use data from the graph to confirm this statement.

[ 2 ]

(b)

Three separate experiments were carried out at 400 K with different starting concentrations of $\mathrm{H}_{2}(\mathrm{~g})$ and $\mathrm{I}_{2}(\mathrm{~g})$ . The results are shown in the table.

[ 6 ]

(i)

Use the data, and the order of reaction with respect to $\mathrm{H}_{2}(\mathrm{~g})$ given in (a)(ii), to deduce the order of reaction with respect to $\mathrm{I}_{2}(\mathrm{~g})$ .

Explain your answer, giving data in support of your explanation.

[ 3 ]

(ii)

Use information from (a)(ii) and your answer to (b)(i) to write the rate equation for the forward reaction.
rate =

[ 1 ]

(iii)

Use your rate equation and data from experiment 1 to calculate the value of the rate constant, k, for the forward reaction at 400 K . Include units for k.
k= units =

[ 2 ]

[Maximum number: 6]

Fluorine reacts with chlorine dioxide, $\mathrm{ClO}_{2}$ , as shown.

\mathrm{F}_{2}(\mathrm{~g})+2 \mathrm{ClO}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{FClO}_{2}(\mathrm{~g})

The rate of the reaction is first order with respect to the concentration of $F_{2}$ and first order with respect to the concentration of $\mathrm{ClO}_{2}$ . No catalyst is involved.

(a)

(i)

Suggest a two-step mechanism for this reaction.

[ 2 ]

(ii)

Identify the rate-determining step in this mechanism. Explain your answer.

[ 1 ]

(b)

When the rate of the reaction is measured in $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ the numerical value of the rate constant, k, is 1.22 under certain conditions.

(i)

Complete the rate equation for this reaction, stating the overall order of the reaction.

\begin{array}{r} \text { rate }= \\ \text { overall order of reaction }= \end{array}

(ii)

Use your rate equation in (i) to calculate the rate of the reaction when the concentrations of $\mathrm{F}_{2}$ and $\mathrm{ClO}_{2}$ are both $2.00 \times 10^{-3} \mathrm{moldm}^{-3}$ .
rate = $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

(c)

Under different conditions, and in the presence of a large excess of $\mathrm{ClO}_{2}$ , the rate equation is as shown.

\text { rate }=k_{1}\left[F_{2}\right]

The half-life, $t_{\frac{1}{2}}$ , of the concentration of $F_{2}$ is 4.00 s under these conditions.

[ 3 ]

(i)

Calculate the numerical value of $k_{1}$ , giving its units.

Give your answer to three significant figures.

k_{1}=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \text { units }

[ 2 ]

(ii)

An experiment is performed under these conditions in which the starting concentration of $F_{2}$ is $0.00200 \mathrm{moldm}^{-3}$ .

Draw a graph on the grid in Fig. 1.1 to show how the concentration of $F_{2}$ changes over the first 12s of the reaction.

Fig. 1.1

[ 1 ]

(iii)

Use your graph in Fig. 1.1 to find the rate of the reaction when the concentration of $F_{2}$ is $0.00100 \mathrm{~mol} \mathrm{dm}^{-3}$ . Show your working on the graph.
rate = moldm $\mathrm{m}^{-3} \mathrm{~s}^{-1}$

[Maximum number: 7]

Propanone, $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ , reacts with iodine, $\mathrm{I}_{2}$ , in the presence of an acid catalyst.

\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}+\mathrm{H}^{+}+\mathrm{I}^{-}

The rate equation for this reaction is shown.

\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{H}^{+}\right]

(a)

Complete Table 1.1 to describe the order of the reaction.

Table 1.1

[ 2 ]

(b)

An experiment is performed using a large excess of $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ and a large excess of $\mathrm{H}^{+}$ (aq). The initial concentration of $\mathrm{I}_{2}$ is $1.00 \times 10^{-5} \mathrm{moldm}^{-3}$ . The initial rate of decrease in the $\mathrm{I}_{2}$ concentration is $2.27 \times 10^{-7} \mathrm{moldm}^{-3} \mathrm{~s}^{-1}$ .

[ 2 ]

(i)

Use the axes to draw a graph of $\left[\mathrm{I}_{2}\right]$ against time for the first 10 seconds of the reaction.

[ 1 ]

(ii)

State whether it is possible to calculate the numerical value of the rate constant, k, for this reaction from your graph. Explain your answer.

[ 1 ]

(c)

The experiment is repeated at a different temperature. The initial concentrations of $\mathrm{H}^{+}$ ions, $\mathrm{I}_{2}$ and $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ are all $0.200 \mathrm{~mol} \mathrm{dm}^{-3}$ .

The value of k at this temperature is $2.31 \times 10^{-5} \mathrm{~mol}^{-1} \mathrm{dm}^{3} \mathrm{~s}^{-1}$ .
Calculate the initial rate of this reaction.
rate = $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

(d)

The experiment is repeated using an excess of $\mathrm{H}^{+}(\mathrm{aq})$ . The new rate equation is shown.

\text { rate }=k_{1}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]

[ 1 ]

(i)

The value of $k_{1}$ is $1.1 \times 10^{-3} \mathrm{~s}^{-1}$ . Calculate the value of the half-life, $t_{\frac{1}{2}}$ .

t_{\frac{1}{2}}=

(ii)

Use your answer to (i) to draw a graph of $\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]$ against time for this reaction. The initial value of $\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]$ on your graph should be $0.200 \mathrm{~mol} \mathrm{dm}^{-3}$ . The final value of $\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]$ on your graph should be $0.0250 \mathrm{~mol} \mathrm{dm}^{-3}$ .

time/s

[ 1 ]

(e)

A four-step mechanism is suggested for the overall reaction.

\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}+\mathrm{H}^{+}+\mathrm{I}^{-} \quad \text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{H}^{+}\right]

Part of this mechanism is shown.

step 1: $\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{C}^{+}(\mathrm{OH}) \mathrm{CH}_{3}$

step 2: $\quad \mathrm{CH}_{3} \mathrm{C}^{+}(\mathrm{OH}) \mathrm{CH}_{3} \rightarrow \mathrm{CH}_{3} \mathrm{C}(\mathrm{OH})=\mathrm{CH}_{2}+\mathrm{H}^{+}$

step 3: →

step 4: $\quad \mathrm{CH}_{3} \mathrm{C}^{+}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{I} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}+\mathrm{H}^{+}$

[ 2 ]

(i)

Write an equation for step 3 .

[ 1 ]

(ii)

Suggest the slowest step of the mechanism. Explain your answer.

[ 1 ]

[Maximum number: 6]

Fluorine reacts with chlorine dioxide, $\mathrm{ClO}_{2}$ , as shown.

\mathrm{F}_{2}(\mathrm{~g})+2 \mathrm{ClO}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{FClO}_{2}(\mathrm{~g})

The rate of the reaction is first order with respect to the concentration of $F_{2}$ and first order with respect to the concentration of $\mathrm{ClO}_{2}$ . No catalyst is involved.

(a)

(i)

Suggest a two-step mechanism for this reaction.

[ 2 ]

(ii)

Identify the rate-determining step in this mechanism. Explain your answer.

[ 1 ]

(b)

When the rate of the reaction is measured in $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ the numerical value of the rate constant, k, is 1.22 under certain conditions.

(i)

Complete the rate equation for this reaction, stating the overall order of the reaction.

\begin{array}{r} \text { rate }= \\ \text { overall order of reaction }= \end{array}

(ii)

Use your rate equation in (i) to calculate the rate of the reaction when the concentrations of $\mathrm{F}_{2}$ and $\mathrm{ClO}_{2}$ are both $2.00 \times 10^{-3} \mathrm{moldm}^{-3}$ .
rate = moldm $\mathrm{m}^{-3} \mathrm{~s}^{-1}$

(c)

Under different conditions, and in the presence of a large excess of $\mathrm{ClO}_{2}$ , the rate equation is as shown.

\text { rate }=k_{1}\left[\mathrm{~F}_{2}\right]

The half-life, $t_{\frac{1}{2}}$ , of the concentration of $F_{2}$ is 4.00 s under these conditions.

[ 3 ]

(i)

Calculate the numerical value of $k_{1}$ , giving its units.

Give your answer to three significant figures.

k_{1}=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \text { unit }

[ 2 ]

(ii)

An experiment is performed under these conditions in which the starting concentration of $F_{2}$ is $0.00200 \mathrm{moldm}^{-3}$ .

Draw a graph on the grid in Fig. 1.1 to show how the concentration of $F_{2}$ changes over the first 12s of the reaction.

Fig. 1.1

[ 1 ]

(iii)

Use your graph in Fig. 1.1 to find the rate of the reaction when the concentration of $F_{2}$ is $0.00100 \mathrm{~mol} \mathrm{dm}^{-3}$ . Show your working on the graph.
rate = moldm $\mathrm{m}^{-3} \mathrm{~s}^{-1}$

[Maximum number: 8]

The compound nitrosyl bromide, NOBr, can be formed by the reaction shown.

2 \mathrm{NO}+\mathrm{Br}_{2} \rightleftharpoons 2 \mathrm{NOBr}

(a)

The rate of the reaction was measured at various concentrations of the two reactants, NO and $\mathrm{Br}_{2}$ , and the following results were obtained.

The general form of the rate equation for this reaction is as follows.

\text { rate }=k[\mathrm{NO}]^{a}\left[\mathrm{Br}_{2}\right]^{b}

[ 7 ]

(i)

What is meant by the term order of reaction with respect to a particular reagent?

[ 1 ]

(ii)

Use the data in the table to deduce the values of a and b in the rate equation. Show your reasoning.

[ 2 ]

(iii)

Use the data in the table to calculate the initial rate for experiment 4.

\text { initial rate }=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}

[ 1 ]

(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 ]

(v)

By considering the rate equation, explain why the rate decreases with decreasing temperature.

[ 1 ]

(b)

The reaction between X and Y was studied.

2 X+Y \rightarrow Z

The following sequence of steps is a proposed mechanism for the reaction.

The general form of the rate equation for this reaction is as follows.

\text { rate }=k[\mathrm{X}]^{m}[\mathrm{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 ]

[Maximum number: 8]

The compound chlorine dioxide, $\mathrm{ClO}_{2}$ , can be prepared by the reaction shown.

\mathrm{NaClO}_{2}+\frac{1}{2} \mathrm{Cl}_{2} \rightarrow \mathrm{ClO}_{2}+\mathrm{NaCl}

(a)

The reaction between $\mathrm{ClO}_{2}$ and $\mathrm{F}_{2}$ is shown.

2 \mathrm{ClO}_{2}+\mathrm{F}_{2} \rightarrow 2 \mathrm{ClO}_{2} \mathrm{~F}

The rate of the reaction was measured at various concentrations of the two reactants and the following results were obtained.

The rate equation is rate $=k\left[\mathrm{ClO}_{2}\right]\left[\mathrm{F}_{2}\right]$ .

[ 5 ]

(i)

What is meant by the term order of reaction with respect to a particular reagent?

[ 1 ]

(ii)

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 ]

(iii)

Use the data in the table to calculate
- the initial rate in experiment 2,
initial rate = $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
- $\left[\mathrm{ClO}_{2}\right]$ in experiment 3.

\left[\mathrm{ClO}_{2}\right]=

moldm ${ }^{-3}$

[ 2 ]

(b)

(i)

What is meant by the term rate-determining step?

[ 1 ]

(ii)

The equation for the reaction between $\mathrm{ClO}_{2}$ and $\mathrm{F}_{2}$ is shown.

\begin{gathered} 2 \mathrm{ClO}_{2}+\mathrm{F}_{2} \rightarrow 2 \mathrm{ClO}_{2} \mathrm{~F} \\ \text { rate }=k\left[\mathrm{ClO}_{2}\right]\left[\mathrm{F}_{2}\right] \end{gathered}

The mechanism for this reaction has two steps.
Suggest equations for the two steps of this mechanism, stating which of the two steps is the rate-determining step.
step 1
step 2
rate-determining step =

[ 2 ]

[Maximum number: 9]

Nitrogen monoxide, NO, reacts with oxygen to form nitrogen dioxide, $\mathrm{NO}_{2}$ .

2 \mathrm{NO}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})

The rate equation for the forward reaction is shown.

\text { rate }=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]

(a)

Complete the following table.

[ 1 ]

(b)

Two separate experiments are carried out at $30^{\circ} \mathrm{C}$ to determine the rate of the forward reaction.

[ 3 ]

(i)

Use the data for experiment 1 to calculate the value of the rate constant, k. State the units of k.

k=

units =

[ 2 ]

(ii)

Calculate the value of $[\mathrm{NO}]$ in experiment 2 .

[NO] =

$\mathrm{mol} \mathrm{dm}^{-3}$

[ 1 ]

(c)

Define the term rate-determining step.

[ 1 ]

(d)

Peroxodisulfate ions, $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}$ , react with iodide ions, $\mathrm{I}^{-}$ .

\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{I}^{-} \rightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}_{2}

The rate equation for the reaction in the absence of any catalyst is shown.

\text { rate }=k\left[\mathrm{~S}_{2} \mathrm{O}_{8}{ }^{2-}\right]\left[\mathrm{I}^{-}\right]

[ 4 ]

(i)

Suggest equations for a two-step mechanism for this reaction, stating which of the two steps is the rate-determining step.
step 1
step 2
rate-determining step =

[ 2 ]

(ii)

A large excess of peroxodisulfate ions is mixed with iodide ions. Immediately after mixing, $\left[\mathrm{I}^{-}\right]=0.00780 \mathrm{~mol} \mathrm{dm}^{-3}$ . Under the conditions used, the half-life of $\left[\mathrm{I}^{-}\right]$ is 48 seconds.

Calculate the iodide ion concentration 192 seconds after the peroxodisulfate and iodide ions are mixed.
iodide ion concentration = $\mathrm{mol} \mathrm{dm}^{-3}$

[ 2 ]

[Maximum number: 7]

Nitrogen monoxide, NO, reacts with oxygen to form nitrogen dioxide, $\mathrm{NO}_{2}$ .

2 \mathrm{NO}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})

The rate equation for the forward reaction is shown.

\text { rate }=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]

(a)

Complete the following table.

[ 1 ]

(b)

Two separate experiments are carried out at $30^{\circ} \mathrm{C}$ to determine the rate of the forward reaction.

[ 3 ]

(i)

Use the data for experiment 1 to calculate the value of the rate constant, k. State the units of k.

k=

units =

[ 2 ]

(ii)

Calculate the value of $[\mathrm{NO}]$ in experiment 2 .

[NO] =

$\mathrm{mol} \mathrm{dm}^{-3}$

[ 1 ]

(c)

Define the term rate-determining step.

[ 1 ]

(d)

Peroxodisulfate ions, $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}$ , react with iodide ions, $\mathrm{I}^{-}$ .

\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{I}^{-} \rightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}_{2}

The rate equation for the reaction in the absence of any catalyst is shown.

\text { rate }=k\left[\mathrm{~S}_{2} \mathrm{O}_{8}{ }^{2-}\right]\left[\mathrm{I}^{-}\right]

[ 2 ]

(i)

Suggest equations for a two-step mechanism for this reaction, stating which of the two steps is the rate-determining step.
step 1
step 2
rate-determining step =

[ 2 ]

(ii)

A large excess of peroxodisulfate ions is mixed with iodide ions. Immediately after mixing, $\left[\mathrm{I}^{-}\right]=0.00780 \mathrm{~mol} \mathrm{dm}^{-3}$ . Under the conditions used, the half-life of $\left[\mathrm{I}^{-}\right]$ is 48 seconds.

Calculate the iodide ion concentration 192 seconds after the peroxodisulfate and iodide ions are mixed.
iodide ion concentration = $\mathrm{mol} \mathrm{dm}^{-3}$