1)   An example of a gas phase chain reaction involving free radicals is hydrodealkylation, in which alkyl aromatics are converted to unsubstituted aromatics. A process for the hydrodealkylation of toluene from petroleum stocks to produce benzene and methane has been developed. The hydrodealkylation occurs in the gas phase at high temperature and involves free radical. The free radical mechanism is believed to proceed by the sequence

Initiation: \( H_2 \xrightarrow{k_1} 2H\cdot \)

Propagation: \[ \begin{cases} H\cdot + C_6H_5CH_3 \xrightarrow{k_2} C_6H_5\cdot + CH_4 \\ C_6H_5\cdot + H_2 \xrightarrow{k_3} C_6H_6 + H\cdot \end{cases} \]

Termination: \( 2H\cdot \xrightarrow{k_4} H_2 \)

The specific reaction rates k₁ and k₄ are defined w.r.t. H₂.

Derive the reaction rate law for the rate of formation of benzene based on this mechanism.

Hint: the reaction is 1/2 order in H₂ and 1st order in toluene.

Solution

Solution

For the elementary equation, we write the rate of formation of benzene as:

\( r_{C_6H_6} = k_3 C_{C_6H_5\cdot} C_{H_2} \)

We need to eliminate the concentration of the free radical \( C_{C_6H_5\cdot} \) by expressing it in terms of the concentrations of toluene and hydrogen.  Under psuedo-steady-state hypothesis, we set the rates of formation of \( C_{H\cdot} \) and \( C_{C_6H_5\cdot} \) equal to zero, i.e.

\( 0 = r_{H\cdot} = 2k_1 C_{H_2} - k_2 C_{H\cdot} C_{C_6H_5CH_3} + k_3 C_{H_2} C_{C_6H_5\cdot} - 2k_4 [C_{H\cdot}]^2 \)

\( 0 = r_{C_6H_5\cdot} = k_2 C_{H\cdot} C_{C_6H_5CH_3} - k_3 C_{H_2} C_{C_6H_5\cdot} \)

Adding equations 2 and 3 yield

\( 0 = r_{H\cdot} = 2k_1 C_{H_2} - 2k_4 [C_{H\cdot}]^2 \)

Solving for \( C_{H\cdot} \) yields:

\( C_{H\cdot} = \sqrt{\frac{k_1 C_{H_2}}{k_4}} \)

Solving for equation 3 for \( C_{C_6H_5\cdot} \) yields

\( C_{C_6H_5\cdot} = \frac{k_2 C_{C_6H_5CH_3} \sqrt{\frac{k_1 C_{H_2}}{k_4}}}{k_3 C_{H_2}} \)

We can now eliminate \( C_{C_6H_5\cdot} \) from the reaction rate expression (1) to obtain

\( r_{C_6H_6} = k_2 C_{C_6H_5CH_3} \sqrt{\frac{k_1 C_{H_2}}{k_4}} \)

\( r_{C_6H_6} = k (C_{H_2})^{1/2} (C_T) \)

\( k = 10^{10.5} \exp\left(\frac{-50,000}{RT}\right) \)

\( C_T = \text{concentration of toluene} \)

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