The equation for the iodination of acetone in acidic solution is $$\mathrm{CH}_{3} \mathrm{COCH}_{3}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q)$$ The rate of the reaction is found to be dependent not only on the concentration of the reactants but also on the hydrogen ion concentration. Hence the rate expression of this reaction is $$\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]^{m}\left[\mathrm{I}_{2}\right]^{n}\left[\mathrm{H}^{+}\right]^{p}$$ The rate is obtained by following the disappearance of iodine using starch as an indicator. The following data are obtained: $$ \begin{array}{cccc} \hline\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right] & \left.\mathrm{[H}^{+}\right] & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 0.80 & 0.20 & 0.001 & 4.2 \times 10^{-6} \\ 1.6 & 0.20 & 0.001 & 8.2 \times 10^{-6} \\ 0.80 & 0.40 & 0.001 & 8.7 \times 10^{-6} \\ 0.80 & 0.20 & 0.0005 & 4.3 \times 10^{-6} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to each reactant? (b) Write the rate expression for the reaction. (c) Calculate \(k\). (d) What is the rate of the reaction when \(\left[\mathrm{H}^{+}\right]=0.933 M\) and \(\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]=3\left[\mathrm{H}^{+}\right]=10\left[\mathrm{I}^{-}\right] ?\)

Step by step solution

01

(a) Determine the order of the reaction with respect to each reactant

To determine the order of the reaction with respect to each reactant, we need to examine how the initial rate changes when the concentration of a single reactant is changed. Let's compare experiments 1 & 2 and experiments 1 & 3. For CH3COCH3: Experiments 1 & 2: $$\frac{(\text{Rate 2})}{(\text{Rate 1})} = \frac{8.2 \times 10^{-6}}{4.2 \times 10^{-6}} = 1.952$$ $$\frac{\left[\mathrm{CH}_{3}\mathrm{COCH}_{3}\right]_{2}}{\left[\mathrm{CH}_{3}\mathrm{COCH}_{3}\right]_{1}} = \frac{1.6}{0.8} = 2$$ Since the ratio of the rates is approximately equal to the ratio of the concentrations, the order \(m\) of the reaction with respect to CH3COCH3 is 1. For H+: Experiments 1 & 3: $$\frac{(\text{Rate 3})}{(\text{Rate 1})} = \frac{8.7 \times 10^{-6}}{4.2 \times 10^{-6}} = 2.071$$ $$\frac{\left[H^{+}\right]_{3}}{\left[H^{+}\right]_{1}} = \frac{0.4}{0.2} = 2$$ Since the ratio of the rates is approximately equal to the ratio of the concentrations, the order \(p\) of the reaction with respect to H+ is 1. For I2: Experiments 1 & 4: $$\frac{(\text{Rate 4})}{(\text{Rate 1})} = \frac{4.3 \times 10^{-6}}{4.2 \times 10^{-6}} = 1.024$$ $$\frac{\left[I_2\right]_{4}}{\left[I_2\right]_{1}} = \frac{0.0005}{0.001} = 0.5$$ Since the ratio of the rates is approximately equal to the square root of the ratio of the concentrations, the order \(n\) of the reaction with respect to I2 is 2.

02

(b) Write the rate expression for the reaction

Now that we have determined the order of the reaction with respect to each reactant, we can write the rate expression: $$\text{rate} = k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]^{1}\left[\mathrm{I}_{2}\right]^{2}\left[\mathrm{H}^{+}\right]^{1}$$

03

(c) Calculate \(k\)

We will use the data from experiment 1 to calculate the rate constant k. $$\text{rate} = k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{I}_{2}\right]^{2}\left[\mathrm{H}^{+}\right]$$ $$k = \frac{\text{rate}}{\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{I}_{2}\right]^{2}\left[\mathrm{H}^{+}\right]} = \frac{4.2 \times 10^{-6}}{(0.80)(0.001)^{2}(0.20)} = 2.625 \times 10^{-2} M^{-2}s^{-1}$$ So, the rate constant \(k = 2.625 \times 10^{-2} M^{-2}s^{-1}\).

04

(d) What is the rate of the reaction under given conditions?

To find the rate of the reaction under given conditions, we must use the rate expression obtained in step (b) and the provided concentrations: \(\left[H^{+}\right]=0.933 M\), \(\left[\mathrm{CH}_{3}\mathrm{COCH}_{3}\right]=3\left[H^{+}\right]=2.799 M\), and \(\left[I^{-}\right]=0.1\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right] = 0.2799 M\). Since I− is a product, its concentration won't affect the rate. The rate expression can be written as: $$\text{rate} = k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{I}_{2}\right]^{2}\left[\mathrm{H}^{+}\right]$$ $$\text{rate} = (2.625 \times 10^{-2} M^{-2}s^{-1})(2.799 M)(0.2799 M)^{2}(0.933 M) = 1.98 \times 10^{-2} M.s^{-1}$$ So, the rate of the reaction under given conditions is \(1.98 \times 10^{-2} M.s^{-1}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!