IChO 2003 · Theory problem 35

Kinetics

The acid-catalyzed reaction \(\ce{CH3COCH3 + I2 -> CH3COCH2I + HI}\) was found to be of first order with respect to hydrogen ions. At constant hydrogen ion concentration the time needed for the concentration of iodine to be reduced by \(0.010\ \mathrm{mol\,dm^{-3}}\) was measured under various concentrations of the reactants.

Based on the information provided in the table, answer fulfil the following tasks:

\([\ce{CH3COCH3}]\) \([\ce{I2}]\) Time
\((\mathrm{mol\,dm^{-3}})\) \((\mathrm{mol\,dm^{-3}})\) \((\mathrm{min})\)
0.25 0.050 7.2
0.50 0.050 3.6
1.00 0.050 1.8
0.50 0.100 3.6
0.25 0.100 …
1.50 … …
… … 0.36

a) Derive the rate law for the reaction and calculate the rate constant.

b) Calculate the time needed for 75 % of \(\ce{CH3COCH3}\) to react in excess \(\ce{I2}\).

c) Show graphically the dependence of the rate on \([\ce{CH3COCH3}]\) and on \([\ce{I2}]\), for fixed initial concentration of the other reagents.

d) If the rate is doubled by raising the temperature by 10 °C from 298 K, calculate the activation energy for this reaction.

Official English text of the problem.

Answers

  • a) The rate law is \[ v=k[\ce{H+}][\ce{CH3COCH3}]. \] At the fixed hydrogen-ion concentration, \[ \boxed{k_{\mathrm{obs}}\equiv k[\ce{H+}] \simeq5.6\times10^{-3}\ \mathrm{min^{-1}}.} \] The full constant \(k\) cannot be evaluated numerically without \([\ce{H+}]\). The missing times are 7.2 min and 1.2 min; the missing acetone concentration is \(5.0\ \mathrm{mol\,dm^{-3}}\). The missing iodine concentrations are not uniquely determined.

  • b) At the same constant hydrogen-ion concentration, \[ \boxed{t_{75\%}\simeq2.5\times10^2\ \mathrm{min}\simeq4.2\ \mathrm{h}.} \]

  • c) Rate versus acetone concentration: a straight line through the origin, with slope \(k_{\mathrm{obs}}\). Rate versus iodine concentration: a horizontal line over the range where the zero-order iodine dependence applies.

  • d) \[ \boxed{E_{\mathrm a}\simeq53\ \mathrm{kJ\,mol^{-1}}.} \]

Solution

a) Rate law and rate constant

The species are acetone (propan-2-one), CC(=O)C; molecular iodine, II; iodoacetone (1-iodopropan-2-one), CC(=O)CI; hydrogen iodide, [H]I; and the hydrogen-ion catalyst, [H+].

The balanced reaction is \[ \ce{CH3COCH3 + I2 -> CH3COCH2I + HI}. \] Its \(1:1\) stoichiometry gives \[ v=-\frac{d[\ce{I2}]}{dt} =-\frac{d[\ce{CH3COCH3}]}{dt}. \]

Since first-order dependence on hydrogen ions is given, write \[ v=k[\ce{H+}][\ce{CH3COCH3}]^m[\ce{I2}]^n. \]

Each measurement corresponds to the same iodine concentration decrease, so the average rate is \[ \bar v=\frac{0.010\ \mathrm{mol\,dm^{-3}}}{t}. \] The acetone concentration changes by only \(1\)–\(4\%\) in the first four experiments, so these average rates approximate the initial rates.

At fixed iodine concentration, doubling the acetone concentration halves the time: \[ \frac{v_2}{v_1}=\frac{7.2}{3.6}=2 =\left(\frac{0.50}{0.25}\right)^m. \] Thus \(m=1\), also confirmed by the next doubling from \(0.50\) to \(1.00\ \mathrm{mol\,dm^{-3}}\).

At fixed acetone concentration, doubling the iodine concentration leaves the time unchanged: \[ \frac{v_4}{v_2}=\frac{3.6}{3.6}=1 =\left(\frac{0.100}{0.050}\right)^n. \] Thus \(n=0\).

Therefore, \[ \boxed{v=k[\ce{H+}][\ce{CH3COCH3}].} \] The reaction is first order in acetone, first order in hydrogen ions, and zero order in iodine: second order overall.

At constant hydrogen-ion concentration, define \[ k_{\mathrm{obs}}=k[\ce{H+}], \qquad v=k_{\mathrm{obs}}[\ce{CH3COCH3}]. \] Using the second experiment, \[ k_{\mathrm{obs}} \simeq \frac{0.010\ \mathrm{mol\,dm^{-3}}} {(3.6\ \mathrm{min})(0.50\ \mathrm{mol\,dm^{-3}})} =5.56\times10^{-3}\ \mathrm{min^{-1}}. \]

The full rate constant is \[ k=\frac{k_{\mathrm{obs}}}{[\ce{H+}]}, \] with units \(\mathrm{dm^3\,mol^{-1}\,min^{-1}}\). Its numerical value requires the hydrogen-ion concentration.

Under the same short-interval approximation, \[ [\ce{CH3COCH3}]\,t =\frac{0.010\ \mathrm{mol\,dm^{-3}}}{k_{\mathrm{obs}}} =1.8\ \mathrm{mol\,dm^{-3}\,min}. \] Hence the incomplete rows become:

\([\ce{CH3COCH3}]\ /\ \mathrm{mol\,dm^{-3}}\) \([\ce{I2}]\ /\ \mathrm{mol\,dm^{-3}}\) Time / min
0.25 0.100 7.2
1.50 Not determined 1.2
5.0 Not determined 0.36

The unspecified iodine concentrations must permit the stated concentration decrease and lie within the regime of zero-order iodine dependence.

b) Time for 75% conversion

With iodine in excess and hydrogen-ion concentration held constant, \[ -\frac{d[\ce{CH3COCH3}]}{dt} =k_{\mathrm{obs}}[\ce{CH3COCH3}]. \] Integration gives \[ \ln\!\left( \frac{[\ce{CH3COCH3}]_0}{[\ce{CH3COCH3}]_t} \right)=k_{\mathrm{obs}}t. \]

After \(75\%\) has reacted, \(25\%\) remains: \[ [\ce{CH3COCH3}]_t=0.25[\ce{CH3COCH3}]_0. \] Therefore, \[ t_{75\%} =\frac{\ln4}{k_{\mathrm{obs}}} =\frac{1.3863}{5.5556\times10^{-3}\ \mathrm{min^{-1}}} =249.5\ \mathrm{min}. \] Thus \[ \boxed{t_{75\%}\simeq250\ \mathrm{min}\simeq4.2\ \mathrm{h}.} \]

c) Dependence of rate on reactant concentrations

For both plots, put concentration in \(\mathrm{mol\,dm^{-3}}\) on the horizontal axis and rate in \(\mathrm{mol\,dm^{-3}\,min^{-1}}\) on the vertical axis.

  • Rate against acetone concentration: keeping hydrogen-ion and iodine concentrations fixed, \[ v=k_{\mathrm{obs}}[\ce{CH3COCH3}]. \] Draw a straight line through the origin with slope \(5.6\times10^{-3}\ \mathrm{min^{-1}}\). Representative points are \[ (0.25,\ 1.39\times10^{-3}),\quad (0.50,\ 2.78\times10^{-3}),\quad (1.00,\ 5.56\times10^{-3}). \]

  • Rate against iodine concentration: keeping hydrogen-ion and acetone concentrations fixed, \[ v=k_{\mathrm{obs}}[\ce{CH3COCH3}]_{\mathrm{fixed}}. \] Draw a horizontal line. For example, at an acetone concentration of \(0.50\ \mathrm{mol\,dm^{-3}}\), its height is \(2.78\times10^{-3}\ \mathrm{mol\,dm^{-3}\,min^{-1}}\), including at iodine concentrations \(0.050\) and \(0.100\ \mathrm{mol\,dm^{-3}}\). This relation does not apply after iodine is exhausted.

d) Activation energy

At unchanged reactant concentrations, doubling the rate means doubling the rate constant. The temperatures are \[ T_1=298\ \mathrm K,\qquad T_2=308\ \mathrm K. \]

The Arrhenius relation gives \[ \ln\!\left(\frac{k_2}{k_1}\right) =\frac{E_{\mathrm a}}{R} \left(\frac1{T_1}-\frac1{T_2}\right). \] Using \(k_2/k_1=2\) and \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\), \[ E_{\mathrm a} = \frac{(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})\ln2} {(1/298-1/308)\ \mathrm{K^{-1}}} =5.289\times10^4\ \mathrm{J\,mol^{-1}}. \] Therefore, \[ \boxed{E_{\mathrm a}\simeq53\ \mathrm{kJ\,mol^{-1}}.} \]

The solution was checked against the official solution.