Consider the following reaction of methane with molecular chlorine:CH4(g)+Cl2(g)→CH3CHcl(g)+HCl(g)Experimental studies have shown that the rate law for this reaction is one-half order with respect to Cl2. Is the following mechanism consistent with this behavior? Cl2→2Cl·Cl·+CH4→HCl+CH3·CH3·+Cl2→

Question

Consider the following reaction of methane with molecular chlorine:

[latex]CH_{4}\left( g ight)+Cl_{2}\left( g ight) o CH_{3}Cl\left( g ight)+HCl\left( g ight)[/latex]

Experimental studies have shown that the rate law for this reaction is one-half order with respect to [latex]Cl_{2}[/latex]. Is the following mechanism consistent with this behavior?

[latex]Cl_{2} \overset{k_{1}}\longrightarrow 2 Cl\cdot [/latex]

[latex]Cl\cdot +CH_{4}\overset{k_{2}}\longrightarrow HCl+CH_{3}\cdot [/latex]

[latex]CH_{3}\cdot+Cl_{2} \overset{k_{3}}\longrightarrow CH_{3}Cl+Cl\cdot [/latex]

[latex]Cl\cdot +Cl\cdot \overset{k_{4}}\longrightarrow Cl_{2}[/latex]

Accepted Answer

The rate of reaction in terms of product [latex]HCl[/latex] is given by

[latex]R=\frac{d\left[ HCl \right]}{dt}=k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right][/latex]

Because [latex]Cl\cdot [/latex] is a reaction intermediate, it cannot appear in the final rate law expression, and [latex]Cl\cdot [/latex] must be expressed in terms of [latex]\left[ CH_{4} \right][/latex] and [latex]\left[ Cl_{2} \right][/latex]. The differential rate expressions for [latex]\left[ Cl\cdot \right][/latex] and [latex]\left[ CH_{3}Cl\cdot \right][/latex] are

[latex]\frac{d\left[ Cl\cdot \right]}{dt}=2k_{1}\left[ Cl_{2} \right]-k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]+k_{3}\left[ CH_{3}\cdot \right]\left[ Cl_{2} \right]-2k_{4}\left[ Cl\cdot \right]^{2}[/latex]

[latex]\frac{d\left[ CH_{3}\cdot\right]}{dt}=k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]-k_{3}\left[ CH_{3}\cdot \right]\left[ Cl_{2} \right][/latex]

Applying the steady-state approximation to the expression for [latex]\left[ CH_{3}\cdot\right][/latex] yields

[latex]\left[ CH_{3}\cdot\right] =\frac{k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]}{k_{3}\left[ Cl_{2} \right]}[/latex]

Next, we substitute this definition of [latex]\left[ CH_{3}\cdot\right][/latex] into the differential rate expression for [latex]\left[ Cl\cdot \right][/latex] and apply the steady state approximation:

[latex]0=2k_{1}\left[ Cl_{2} \right]-k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]+k_{3}\left[ CH_{3}\cdot \right]\left[ Cl_{2} \right]-2k_{4}\left[ Cl\cdot \right]^{2}[/latex]

[latex]0=2k_{1}\left[ Cl_{2} \right]-k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]+k_{3}\left( \frac{k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]}{k_{3}\left[ Cl_{2} \right]} \right)\left[ Cl_{2} \right]-2k_{4}\left[ Cl\cdot \right]^{2}[/latex]

[latex]0=2k_{1}\left[ Cl_{2} \right]-2k_{4}\left[ Cl\cdot \right]^{2}[/latex]

[latex]\left[ Cl\cdot \right]=\left( \frac{k_{1}}{k_{4}} \left[ Cl_{2} \right]\right)^{1/2}[/latex]

With this result, the predicted rate law expression becomes

[latex]R=k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]=k_{2}\left( \frac{k_{1}}{k_{4}} \right)^{1/2}\left[ CH_{4} \right]\left[ Cl_{2} \right]^{1/2}[/latex]

The mechanism is consistent with the experimentally observed one-half order dependence on [latex]\left[ Cl_{2} \right][/latex].

Question 19.3: Consider the following reaction of methane with molecular ch...

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