Question 19.3: Consider the following reaction of methane with molecular ch...
Consider the following reaction of methane with molecular chlorine:
CH_{4}\left( g \right)+Cl_{2}\left( g \right)\to CH_{3}Cl\left( g \right)+HCl\left( g \right)Experimental studies have shown that the rate law for this reaction is one-half order with respect to Cl_{2}. Is the following mechanism consistent with this behavior?
Cl_{2} \overset{k_{1}}\longrightarrow 2 Cl\cdotCl\cdot +CH_{4}\overset{k_{2}}\longrightarrow HCl+CH_{3}\cdot
CH_{3}\cdot+Cl_{2} \overset{k_{3}}\longrightarrow CH_{3}Cl+Cl\cdot
Cl\cdot +Cl\cdot \overset{k_{4}}\longrightarrow Cl_{2}
The rate of reaction in terms of product HCl is given by
R=\frac{d\left[ HCl \right]}{dt}=k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]Because Cl\cdot is a reaction intermediate, it cannot appear in the final rate law expression, and Cl\cdot must be expressed in terms of \left[ CH_{4} \right] and \left[ Cl_{2} \right]. The differential rate expressions for \left[ Cl\cdot \right] and \left[ CH_{3}Cl\cdot \right] are
\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}\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]
Applying the steady-state approximation to the expression for \left[ CH_{3}\cdot\right] yields
\left[ CH_{3}\cdot\right] =\frac{k_{2}\left[ Cl\cdot \right]\left[ CH_{4} \right]}{k_{3}\left[ Cl_{2} \right]}Next, we substitute this definition of \left[ CH_{3}\cdot\right] into the differential rate expression for \left[ Cl\cdot \right] and apply the steady state approximation:
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}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}
0=2k_{1}\left[ Cl_{2} \right]-2k_{4}\left[ Cl\cdot \right]^{2}
\left[ Cl\cdot \right]=\left( \frac{k_{1}}{k_{4}} \left[ Cl_{2} \right]\right)^{1/2}
With this result, the predicted rate law expression becomes
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}The mechanism is consistent with the experimentally observed one-half order dependence on \left[ Cl_{2} \right].