IChO 1981 · Theory problem 6

Kinetics of the catalytic decomposition of isopropanol on V2O5

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The catalytic decomposition of isopropanol on the surface of a \(\ce{V2O5}\) catalyst, leading to the products in the scheme, satisfies a first order kinetic equation.

Figure 1

Five seconds after initiation of the reaction at 590 K, the concentrations of the components in the reaction mixture are:

\[ \begin{aligned} c_{\mathrm A}&=28.2\ \mathrm{mmol\,dm^{-3}}\\ c_{\mathrm B}&=7.8\ \mathrm{mmol\,dm^{-3}}\\ c_{\mathrm C}&=8.3\ \mathrm{mmol\,dm^{-3}}\\ c_{\mathrm D}&=1.8\ \mathrm{mmol\,dm^{-3}} \end{aligned} \]

6.1 What is the initial concentration \(c_0\) of \(\ce{C3H7OH}\) in the system?

6.2 What is the value of the rate constant \(k\) for the process:

Figure 2

6.3 What is the interval of time \((\tau_{1/2})\) in which the concentration of \(\ce{C3H7OH}\) will reach the value \(c=c_0/2\)?

6.4 What are the values of rate constants \(k_1\), \(k_2\), and \(k_3\)?

6.5 What are the values of concentrations \(c_{\mathrm B}\), \(c_{\mathrm C}\), \(c_{\mathrm D}\) at \(t=\tau_{1/2}\)?

The equation describing the concentration changes of A with time \(t\) for the first order reaction has the form:

\[ c_{\mathrm A}=c_0\exp(-kt) \]

or

\[ \log(c_0/c_{\mathrm A})=0.4343\,kt \]

or

\[ \ln(c_0/c_{\mathrm A})=kt \]

Fill in the table with the answers obtained.

1 \(c_0=\)
2 \(k=\)
3 \(\tau_{1/2}\)
4 \(k_1=\)
\(k_2=\)
\(k_3=\)
5 \(c_{\mathrm B}=\)
\(c_{\mathrm C}=\)
\(c_{\mathrm D}=\)

Official English text of the problem.

Answers

  • 6.1: \(c_0=\boxed{46.1\ \mathrm{mmol\,dm^{-3}}}\).
  • 6.2: \(k=\boxed{9.83\times10^{-2}\ \mathrm{s^{-1}}}\).
  • 6.3: \(\tau_{1/2}=\boxed{7.05\ \mathrm{s}}\), measured from initiation.
  • 6.4: \(\boxed{k_1=4.56\times10^{-2}\ \mathrm{s^{-1}}}\), \(\boxed{k_2=4.28\times10^{-2}\ \mathrm{s^{-1}}}\), \(\boxed{k_3=9.88\times10^{-3}\ \mathrm{s^{-1}}}\).
  • 6.5: At \(t=\tau_{1/2}\), \(\boxed{c_{\mathrm B}=10.0\ \mathrm{mmol\,dm^{-3}}}\), \(\boxed{c_{\mathrm C}=10.7\ \mathrm{mmol\,dm^{-3}}}\), and \(\boxed{c_{\mathrm D}=2.32\ \mathrm{mmol\,dm^{-3}}}\).

Solution

6.1 — Initial concentration

The substrate \(A\) is isopropanol (propan-2-ol), CC(O)C. Dehydrogenation of this secondary alcohol gives the ketone \(B\), acetone (propan-2-one), CC(=O)C; dehydration gives the alkene \(C\), propene, C=CC. The saturated product \(D\), \(\ce{C3H8}\), is propane, CCC.

The stated kinetic scheme consists of three parallel, irreversible pathways. Initially the products are absent. Each molecule of \(A\) consumed yields one molecule of \(B\), \(C\), or \(D\), retaining all three carbon atoms. Thus, at constant volume,

\[ c_0=c_{\mathrm A}+c_{\mathrm B}+c_{\mathrm C}+c_{\mathrm D}. \]

Using the concentrations at \(t=5\ \mathrm{s}\),

\[ c_0=(28.2+7.8+8.3+1.8)\ \mathrm{mmol\,dm^{-3}} =\boxed{46.1\ \mathrm{mmol\,dm^{-3}}}. \]

6.2 — Overall rate constant

The three pathways contribute additively to the disappearance of \(A\):

\[ -\frac{dc_{\mathrm A}}{dt} =(k_1+k_2+k_3)c_{\mathrm A} =kc_{\mathrm A}. \]

Therefore,

\[ k=k_1+k_2+k_3. \]

The integrated first-order equation gives

\[ k=\frac{1}{t}\ln\!\left(\frac{c_0}{c_{\mathrm A}}\right). \]

Hence, at \(590\ \mathrm K\),

\[ k=\frac{1}{5\ \mathrm{s}} \ln\!\left(\frac{46.1}{28.2}\right) =0.0982982\ \mathrm{s^{-1}} \approx\boxed{9.83\times10^{-2}\ \mathrm{s^{-1}}}. \]

6.3 — Half-life

At the half-life, \(c_{\mathrm A}=c_0/2\), so

\[ \frac{c_0}{2}=c_0e^{-k\tau_{1/2}}. \]

Consequently,

\[ \tau_{1/2}=\frac{\ln 2}{k} =\frac{0.693147}{0.0982982\ \mathrm{s^{-1}}} =\boxed{7.05\ \mathrm{s}}. \]

This is the elapsed time from the start of the reaction.

6.4 — Individual rate constants

The diagram assigns \(k_2\) to \(B\), \(k_1\) to \(C\), and \(k_3\) to \(D\). Their formation rates are therefore

\[ \frac{dc_{\mathrm B}}{dt}=k_2c_{\mathrm A}, \qquad \frac{dc_{\mathrm C}}{dt}=k_1c_{\mathrm A}, \qquad \frac{dc_{\mathrm D}}{dt}=k_3c_{\mathrm A}. \]

For example, integrating the first equation with \(c_{\mathrm B}(0)=0\) gives

\[ c_{\mathrm B}(t) =k_2\int_0^t c_0e^{-kt'}\,dt' =\frac{k_2}{k}c_0(1-e^{-kt}) =\frac{k_2}{k}(c_0-c_{\mathrm A}). \]

Similarly,

\[ c_{\mathrm C}=\frac{k_1}{k}(c_0-c_{\mathrm A}), \qquad c_{\mathrm D}=\frac{k_3}{k}(c_0-c_{\mathrm A}). \]

The concentration of \(A\) consumed after five seconds is

\[ c_0-c_{\mathrm A}=46.1-28.2 =17.9\ \mathrm{mmol\,dm^{-3}}. \]

Thus,

\[ \begin{aligned} k_1 &=k\frac{c_{\mathrm C}}{c_0-c_{\mathrm A}} =(0.0982982)\frac{8.3}{17.9}\ \mathrm{s^{-1}} \approx\boxed{4.56\times10^{-2}\ \mathrm{s^{-1}}},\\[4pt] k_2 &=k\frac{c_{\mathrm B}}{c_0-c_{\mathrm A}} =(0.0982982)\frac{7.8}{17.9}\ \mathrm{s^{-1}} \approx\boxed{4.28\times10^{-2}\ \mathrm{s^{-1}}},\\[4pt] k_3 &=k\frac{c_{\mathrm D}}{c_0-c_{\mathrm A}} =(0.0982982)\frac{1.8}{17.9}\ \mathrm{s^{-1}} \approx\boxed{9.88\times10^{-3}\ \mathrm{s^{-1}}}. \end{aligned} \]

The unrounded values satisfy \(k_1+k_2+k_3=k\).

6.5 — Product concentrations at the half-life

At \(t=\tau_{1/2}\), the concentration of \(A\) consumed is

\[ c_0-c_{\mathrm A}=\frac{c_0}{2} =23.05\ \mathrm{mmol\,dm^{-3}}. \]

Because all three products form directly from the same reactant by first-order pathways, their proportions remain constant:

\[ c_{\mathrm B}:c_{\mathrm C}:c_{\mathrm D} =k_2:k_1:k_3 =7.8:8.3:1.8. \]

Therefore,

\[ \begin{aligned} c_{\mathrm B}(\tau_{1/2}) &=\frac{7.8}{17.9}(23.05) =10.0441\ \mathrm{mmol\,dm^{-3}} \approx\boxed{10.0\ \mathrm{mmol\,dm^{-3}}},\\[4pt] c_{\mathrm C}(\tau_{1/2}) &=\frac{8.3}{17.9}(23.05) =10.6880\ \mathrm{mmol\,dm^{-3}} \approx\boxed{10.7\ \mathrm{mmol\,dm^{-3}}},\\[4pt] c_{\mathrm D}(\tau_{1/2}) &=\frac{1.8}{17.9}(23.05) =2.31788\ \mathrm{mmol\,dm^{-3}} \approx\boxed{2.32\ \mathrm{mmol\,dm^{-3}}}. \end{aligned} \]

The unrounded product concentrations sum to \(23.05\ \mathrm{mmol\,dm^{-3}}=c_0/2\), as required.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.