Chemistry - Chemical Kinetics Question with Solution | TestHub

The overall rate law is given as:

rate = $7.1\times 10^{-5}$ [tert-butyl bromide][HO⁻] + $1.5\times 10^{-5}$ [tert-butyl bromide]

The first term corresponds to the E2 pathway:

Rate(E2) = $k_2$ [tert-butyl bromide][HO⁻] = $7.1\times 10^{-5}$ [tert-butyl bromide][HO⁻]

The second term corresponds to the E1 pathway:

Rate(E1) = $k_1$ [tert-butyl bromide] = $1.5\times 10^{-5}$ [tert-butyl bromide]

We are given [HO⁻] = 5.0 M.

Let [tert-butyl bromide] = X for simplicity.

Rate(E2) = $7.1\times 10^{-5}\times X\times 5.0=35.5\times 10^{-5}X$

Rate(E1) = $1.5\times 10^{-5}X$

Total Rate = Rate(E2) + Rate(E1) = $35.5\times 10^{-5}X+1.5\times 10^{-5}X=37.0\times 10^{-5}X$

The percentage of the reaction that takes place by the E2 pathway is:

$$\frac{Rate(E2)}{TotalRate}\times 100\%$$

$$\frac{35.5\times 10^{-5}X}{37.0\times 10^{-5}X}\times 100\%$$

$$\frac{35.5}{37.0}\times 100\%\approx 0.9594\times 100\%\approx 96\%$$