Question

HCl reacts with propene (CH₃CHCH₂) in the gas phase according to the overall reaction.

HCl + CH₃CH CH₂ → CH₃CHClCH₃

The experimental rate expression is

rate = k[ HCl ]₃ [ CH₃CHCH₂ ]

Which, if any, of the following mechanisms are consistent with the observed rate expression?

(a) HCl + HCl ⇄ H + HCl₂ (Fast)

H + CH₃CH CH₂ → CH₃CH CH₃(Slow)

HCl₂ + CH₃CH CH₃ → CH₃CHClCH₃ + HCl (Fast)

(b) HCl + HCl ⇄ H₂Cl₂

HCl + CH₃CHCH₂ → CH₃CHCl CH₃^*(Slow)

CH₃CHCl CH₃^*+ H₂Cl₂→ CH₃CHCl CH₃+ 2HCl

(C) HCl + CH₃CH CH₂ → H +CH₃CHClCH₂ (Fast equilbrium)

H + HCl ⇄ H₂Cl (Fast equilbrium)

H₂Cl + CH₃CHCl CH₂ → HCl + CH₃CHCl CH₃ (Slow) _,

Short answer

The only rate law that obeys the rate law in the question is b

Step-by-step solution

Step-1:   Determination of rate constant and Rate expression for a :

If the reaction contains several steps,rate determinng step is the slowest step of the reaction.

$\mathrm{HCl}+{\mathrm{CH}}_3\mathrm{CH}{\mathrm{CH}}_2\to {\mathrm{CH}}_3\mathrm{CH}Cl{\mathrm{CH}}_3{\left(\mathrm{Slow}\right)}_{}$

The rate expression for the reaction is

$Rate=k_2\left[H\right]\left[C{H}_{3}CHC{H}_2\right]......\left(1\right)$

Where K₂ is the rate constant of the reaction(slow)

But H is in equilbrium with both HCl and HCl₂ in the first step,so the rate constant for the first step is K₁is then

$$\begin{gathered} K_1=\frac{\left[H\right]\left[HC{l}_2\right]}{{\left[HCl\right]}^2}.......\left(2\right) \\ Accordingtoprincipleof\det ailedbalance \\ {K}_1=\frac{k_1}{k_{-1}}.........\left(3\right) \end{gathered}$$

Where k and $k_{-1}$ are the rate constant value of both forward and reverse reactions.

From equation 2 and 3

$$\begin{gathered} \frac{k_1}{k_{-1}}=\frac{\left[H\right]\left[HC{l}_2\right]}{{\left[HCl\right]}^2}.......\left(2\right) \\ or\left[H\right]=\frac{k_1{\left[HCl\right]}^2}{k_{-1}\left[HC{l}_2\right]}.......\left(3\right) \\ Put\left[H\right]valueinequation-1 \\ Rate=\frac{k_2{k}_1{\left[HCl\right]}^2\left[C{H}_{3}CHC{H}_2\right]}{k_{-1}\left[HC{l}_2\right]}......\left(4\right) \end{gathered}$$

The given rate expression does not match the rate law given in the question

Step-2:  Determination of rate constant and Rate expression of b:

b) If the reaction contains several steps ,slowest step is the rate determining step.

$HCl+C{H}_{3}CHC{H}_2\to C{H}_{3}CHClC{H}_2.........\left(1\right)$

The rate expession for the above reaction

$Rate=k_3\left[H_{2}C{l}_2\right]\left[C{H}_{3}CHClC{H}_3\right]*.......\left(2\right)$

Where k₃ is the rate constant of third step of the reaction.

It was found that * is in equilbrium with HCl and CH₃CHCH₂ then k₂ of the reaction is

$K_2=\frac{\left[C{H}_{3}CHClC{H}_3\right]*}{\left[HCl\right]\left[C{H}_{3}CHC{H}_2\right]}.......\left(3\right)$

According to principle of detailed balance

$K_2=\frac{k_2}{k_{-2}}.......\left(4\right)$

Where k₂ and k_-2 are the rate constant of forward and backward reaction.

From equation 3 and 4

$\frac{k_2}{k_{-2}}=\frac{\left[C{H}_{3}CHClC{H}_3\right]*}{\left[HCl\right]\left[C{H}_{3}CHC{H}_2\right]}..........\left(5\right)$

Or the concendration$\left[C{H}_{3}CHClC{H}_3\right]*$of is calculated as

$\left[C{H}_{3}CHClC{H}_3\right]*=\frac{k_2\left[HCl\right]\left[C{H}_{3}CHC{H}_2\right]}{k_{-2}}..........\left(6\right)$

Substituting$\left[C{H}_{3}CHClC{H}_3\right]*$in the equation 2

$Rate=\frac{k_2{k}_3\left[H_{2}C{l}_2\right]\left[HCl\right]\left[C{H}_{3}CHC{H}_2\right]}{k_{-2}}.......\left(7\right)$

Again H₂Cl₂ is in equilbrium with HCl in first step,so equilbrium constant K1 for first step of the reaction is given as

$K_1=\frac{\left[H_{2}C{l}_2\right]}{{\left[HCl\right]}^2}.........\left(8\right)$

According to principle of detailed balance

$K_1=\frac{k_1}{k_{-1}}.....\left(9\right)$

Where k₁ and k_-1are the rate constant of forward and back ward reaction.

From the equation 8 and 9

$$\begin{gathered} \frac{k_1}{k_{-1}}=\frac{\left[H_{2}C{l}_2\right]}{{\left[HCl\right]}^2} \\ \left[H_{2}C{l}_2\right]=\frac{k_1{\left[HCl\right]}^2}{k_{-1}}..........\left(10\right) \end{gathered}$$

Substituting$\left[H_{2}C{l}_2\right]$ value in 7 th equation

$$\begin{gathered} Rate=\frac{k_2{k}_3{k}_1{\left[HCl\right]}^3\left[C{H}_{3}CHC{H}_2\right]}{k_{-2}{k}_{-1}}...........\left(11\right) \\ Rate=k{\left[HCl\right]}^3\left[C{H}_{3}CHC{H}_2\right] \\ Wherek=\frac{k_3{k}_2{k}_1}{k_{-2}{k}_{-1}} \end{gathered}$$

Therefore rate law of corresponding mechanism matches the rate law given in the question.

STEP-3: Determination of rate constant and Rate expression:

C) if the reaction contain several steps ,slowest step is the rated determining step

$$\begin{gathered} HCl+C{H}_{3}CHC{H}_2\to C{H}_{3}CHClC{H}_2 \\ Rate\exp ressionforaboveequation \\ Rate=k_3\left[HCl\right]\left[C{H}_{3}CHClC{H}_2\right]......\left(1\right) \end{gathered}$$

$C{H}_{3}CHClC{H}_2$is in equil brium with HCl and $C{H}_{3}CHC{H}_2$ in the equation ,therefore equilbrium constant K₁ is the following

$K_1=\frac{\left[C{H}_{3}CHClC{H}_2\right]\left[H\right]}{\left[HCl\right]\left[C{H}_{3}CHC{H}_2\right]}........\left(2\right)$

According to principle of detailed balance

$K_1=\frac{k_1}{k_{-1}}..........\left(3\right)$

Where k and k_-1 are the rate constants of forward and backward reactions

From equation 3 and 2

Substituting the value of $\left[C{H}_{3}CHClC{H}_2\right]$ in the equation 1

$\left[C{H}_{3}CHClC{H}_2\right]=\frac{k_1\left[HCl\right]\left[C{H}_{3}CHC{H}_2\right]}{k_{-1}\left[H\right]}$

$Rate=\frac{k_3{k}_1\left[H_{2}Cl\right]\left[HCl\right]\left[C{H}_{3}CHC{H}_2\right]}{k_{-1}\left[H\right]}.........\left(4\right)$

Again H₂Cl is equilbrium with HCl ,H in the seacond step ,if the equilbrium constant for seacond step k₂ then

$K_2=\frac{\left[H_{2}Cl\right]}{\left[HCl\right]\left[H\right]}........\left(5\right)$

According to principle of detailed balance

$K_2=\frac{k_2}{k_{-2}}........\left(6\right)$

Where k₂ and k_-2 are rate constant of both forward and backward reaction in the seacond step.

From the equation 5 and 6

$$\begin{gathered} \frac{k_2}{k_{-2}}=\frac{\left[H_{2}Cl\right]}{\left[HCl\right]\left[H\right]}...........\left(7\right) \\ or\left[H_{2}Cl\right]=\frac{k_2\left[HCl\right]\left[H\right]}{k_{-2}}.......\left(8\right) \end{gathered}$$

Substituting the value of $\left[H_{2}Cl\right]$ in the equation 4

$$\begin{gathered} Rate=\frac{k_1{k}_2{k}_3{\left[HCl\right]}^2\left[C{H}_{3}CHC{H}_2\right]}{k_{-2}{k}_{-1}}...........\left(9\right) \\ =k{\left[HCl\right]}^2\left[C{H}_{3}CHC{H}_2\right] \\ k=\frac{k_3{k}_2{k}_1}{k_{-1}{k}_{-2}} \end{gathered}$$

Therefore the rate law of above mechanism does not match the given rate expression given in the question.

The only rate law that obeys the rate law in the question is b