The gas-phase chlorination of dimethyl ether at proceeds according to the overall reaction

CHEMISTRY · Multi-part · A Level · Chemistry · Rate equations · Rate equations

Problem

The gas-phase chlorination of dimethyl ether at

350\,\text{K}

proceeds according to the overall reaction

\mathrm{CH_3OCH_3} + \mathrm{Cl_2} \rightarrow \mathrm{CH_3OCH_2Cl} + \mathrm{HCl}

. The following sequence of elementary steps has been proposed: Step 1:

\mathrm{Cl_2} \rightleftharpoons 2\,\mathrm{Cl}

(fast) Step 2:

\mathrm{Cl} + \mathrm{CH_3OCH_3} \rightarrow \mathrm{CH_3OCH_2} + \mathrm{HCl}

(slow) Step 3:

\mathrm{CH_3OCH_2} + \mathrm{Cl_2} \rightarrow \mathrm{CH_3OCH_2Cl} + \mathrm{Cl}

(fast) Experimental measurements show that, under the conditions studied, the rate is proportional to

[\mathrm{Cl_2}]^{1/2}

and to

[\mathrm{CH_3OCH_3}]^{1}

. Answer the questions that follow. (1) Part 1: (a) Write the rate law predicted by the mechanism in terms of the species present immediately before the rate-limiting step. (2) Part 2: (b) Using your expression from part (a), express the overall rate law only in terms of the reactant concentrations

[\mathrm{Cl_2}]

and

[\mathrm{CH_3OCH_3}]

, matching the exponents given by experiment. (3) Part 3: (c) Under otherwise identical conditions, the initial concentration of

\mathrm{CH_3OCH_3}

is doubled while

[\mathrm{Cl_2}]_0

is kept constant. By what factor does the initial rate change? Explain your reasoning. (4) Part 4: (d) Instead, the initial concentration of

\mathrm{Cl_2}

is quadrupled while

[\mathrm{CH_3OCH_3}]_0

remains the same. Predict the factor by which the initial rate changes, and justify your answer using the reaction orders you obtained.

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