[12%] Iodomethane, CH3I, hydrolyzes irreversibly in dilute solution via two distinct pathways: CH3I( — Chemical Kinetics Chemistry Question
Problem Context
[12%] Iodomethane, CH3I, hydrolyzes irreversibly in dilute solution via two distinct pathways:
CH3I(aq) + H2O(l) → CH3OH(aq) + H+(aq) + I–(aq) Rate = k1[CH3I]
CH3I(aq) + OH–(aq) → CH3OH(aq) + I–(aq) Rate = k2[CH3I][OH–]
A solution is prepared by dissolving enough methyl iodide in pure water to make a 0.0500 M solution. After 1.50 h at 89.9 °C, the concentration of iodide in this solution is measured to be 0.0311 M. What is the value of k1 at 89.9 °C?
Model Answer
If [I–] = 0.0311 M, then [CH3I] = 0.0500 – 0.0311 = 0.0189 M. Since this is a first-order reaction,
ln([CH3I]) = ln([CH3I]0) – k1t
ln(0.0189) = ln(0.0500) – k1(5400 s)
k1 = 1.80 × 10–4 s–1
What would the concentration of iodide be in this solution after 3.00 h at 89.9 °C?
Model Answer
ln([CH3I]) = ln([CH3I]0) – k1t
ln([CH3I]) = ln(0.0500) – (1.80 × 10–4 s–1)(10800 s)
[CH3I] = 7.16 × 10–3 M
[I–] = 0.0500 M – 0.0072 M = 0.0428 M
The hydrolysis of iodomethane is studied in strongly buffered solutions as a function of pH. Under these conditions, the reaction is always found to be first order in iodomethane, with an observed first-order rate constant kobs that varies with pH. The logarithms of kobs (with kobs measured in units of s–1) for the reaction studied at 333 K as a function of pH are shown with filled circles and a solid line on the graph below.
c. What are the values of k1 and k2 for the reaction at 333 K?
Model Answer
Because the reactions take place in a buffer solution, the concentration of hydroxide ion is constant throughout each run, so rate = (k1 + k2[OH–])[CH3I]. Thus, kobs = k1 + k2[OH–]. At low pH, the concentration of hydroxide becomes very small and kobs approaches k1 asymptotically. From the graph, log10(k1) = –5.06, k1 = 8.7 × 10–6 s-1.
The value of k2 can be inferred from any point on the graph at higher pH. An easy value to pick is at pH = 13.0, where log10(kobs) = –3.48, kobs = 3.31 × 10–4 s–1.
kobs = k1 + k2[OH–]
3.31 × 10–4 s–1 = 8.7 × 10–6 s-1 + k2[0.10 M]
k2 = 3.2 × 10–3 M–1 s–1
The analogous data for the reaction studied at 343 K are shown on the graph (filled squares, dashed line). Which has a larger activation energy, reaction 1 or reaction 2? Explain your reasoning.
Model Answer
At pH = 9, hydrolysis is essentially only taking place by reaction 1. The difference in the graphs at this pH, 0.46 log units, indicates that the reaction 1 is faster at 343 K by a factor of 10^0.46 = 2.9. At pH = 13, hydrolysis is taking place almost completely (> 95%) by reaction 2. The difference in the graphs is slightly smaller here, only 0.42 log units corresponding to a factor of 2.6. Reaction 1 therefore has a greater temperature-dependence than reaction 2, and since k = Ae–Ea/RT, reaction 1 must therefore have the greater activation energy.
The value of k2 for CH3CH2I is significantly smaller than the value of k2 for CH3I at a given temperature. Rationalize this difference based on the structures of the transition states for these reactions.
Model Answer
Reaction 2 is bimolecular, with hydroxide ion approaching the carbon 180° away from the iodide in the transition state of the (so-called SN2) reaction. Since the coordination number of carbon increases from 4 to 5 in the transition state, substitution of methyl for hydrogen at the carbon makes it more difficult to attain the transition state due to increased steric crowding. Thus the activation energy is higher for CH3CH2I and the value of k2 correspondingly smaller.