[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 ]

[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 ]

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

(a)

The oxidation of nitrogen(II) oxide is shown in the equation.

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

The initial rate of this reaction was measured, starting with different concentrations of the two reactants. The following results were obtained.

[ 6 ]

(i)

Use the data in the table to determine the order with respect to each reactant. Show your reasoning.

(ii)

Calculate the initial rate in experiment 4. Give your answer to two significant figures.
initial rate = $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

(iii)

Write the rate equation for this reaction.

(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

[ 6 ]

(b)

The compound nitrosyl fluoride, NOF, can be formed by the following reaction.

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

The rate is first order with respect to NO and $\mathrm{F}_{2}$ .
The reaction mechanism has two steps.
Suggest equations for the two steps of this mechanism, stating which is the rate determining slower step.

[ 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: 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 ]