IChO 1990 · Theory problem 5

Chemical kinetics

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Nitramide \(\ce{NO2NH2}\) decomposes slowly in aqueous solution according to the reaction:

Figure 1

The experimental kinetic law is as follows:

\[ \frac{d[\ce{N2O}]}{dt} =k\frac{[\ce{NO2NH2}]}{[\ce{H3O+}]} \]

5.1 What is the apparent order of this reaction in a buffered solution?

5.2 Which of the following mechanisms is the most appropriate for the interpretation of this kinetic law? Justify your answer.

Figure 2
Figure 3
Figure 4

5.3 Show the relationship between the experimentally observed rate constant and the rate constants of the selected mechanism.

5.4 Show that hydroxyl ions catalyze the decomposition of nitramide.

The decomposition is studied in a buffered solution of a constant volume \(V\) at a constant temperature by measuring the partial pressure of the gas \(\ce{N2O}\), considered to be insoluble in water, in a constant volume of the same value \(V\) above the solution. The following results are obtained:

\(t\)(min) 0 5 10 15 20 25
\(p\)(Pa) 0 6800 12400 17200 20800 24000

After a sufficiently long time, the pressure stabilises at 40000 Pa.

5.5 Express the pressure \(p\) as a function of time and the constant \(k'\) such that

\[ k'=\frac{k}{[\ce{H3O+}]}. \]

5.6 Verify graphically that the kinetic law is confirmed by these experimental results.

5.7 Calculate \(k'\) giving its units.

Official English text of the problem.

Answers

  • 5.1: First order, with the constant apparent rate coefficient \(k'=k/[\ce{H3O+}]\).
  • 5.2: Mechanism 3, involving the nitramidate ion [NH-][N+](=O)[O-].
  • 5.3: With elementary mass-action constants for the reactions as written, \[ \boxed{k=\frac{k_4k_5}{k_{-4}}[\ce{H2O}]}. \] If water concentration is absorbed into \(\widetilde{k}_4=k_4[\ce{H2O}]\), then \(k=\widetilde{k}_4k_5/k_{-4}\).
  • 5.4: Hydroxide ions [OH-] accelerate the reaction and are regenerated: \[ v=\frac{k}{K_w}[\ce{NO2NH2}][\ce{OH-}]. \]
  • 5.5: \[ \boxed{p(t)=(40000\ \mathrm{Pa})\left(1-e^{-k't}\right)}. \]
  • 5.6: Plotting \(\ln[p_\infty/(p_\infty-p)]\) against time gives an approximately straight line through the origin, as predicted.
  • 5.7: \[ \boxed{k'\approx3.7\times10^{-2}\ \mathrm{min^{-1}} \approx6.1\times10^{-4}\ \mathrm{s^{-1}}}. \]

Solution

The species used below have the following names and SMILES. No stereochemistry is involved.

Formula Name SMILES
\(\ce{NO2NH2}\) Nitramide N[N+](=O)[O-]
\(\ce{NO2NH-}\) Nitramidate ion, the conjugate base of nitramide [NH-][N+](=O)[O-]
\(\ce{NO2NH3+}\) N-protonated nitramide [NH3+][N+](=O)[O-]
\(\ce{N2O}\) Nitrous oxide, or dinitrogen monoxide N#[N+][O-]
\(\ce{H2O}\) Water O
\(\ce{H3O+}\) Hydronium ion [OH3+]
\(\ce{OH-}\) Hydroxide ion [OH-]

5.1 — Apparent reaction order

A buffer maintains approximately constant \([\ce{H3O+}]\). Consequently,

\[ v=\frac{d[\ce{N2O}]}{dt} =\underbrace{\frac{k}{[\ce{H3O+}]}}_{k'} [\ce{NO2NH2}]. \]

The rate is proportional to the first power of the nitramide concentration. Thus the reaction is apparently first order, or pseudo-first-order, with rate coefficient \(k'\).

5.2 — Selection of the mechanism

For Mechanism 1, the single elementary step is

\[ \ce{NO2NH2 -> N2O + H2O}. \]

It predicts

\[ v=k_1[\ce{NO2NH2}], \]

which does not account for the experimentally observed inverse dependence on \([\ce{H3O+}]\).

For Mechanism 2, the steps are

\[ \ce{NO2NH2 + H3O+ <=> NO2NH3+ + H2O}, \]

\[ \ce{NO2NH3+ -> N2O + H3O+}. \]

The first step is a rapid equilibrium. At constant water concentration,

\[ [\ce{NO2NH3+}] \propto[\ce{NO2NH2}][\ce{H3O+}]. \]

The slow second step therefore gives

\[ v=k_3[\ce{NO2NH3+}] \propto[\ce{NO2NH2}][\ce{H3O+}], \]

predicting a positive first-order dependence on hydronium concentration. This contradicts the observed inverse dependence.

For Mechanism 3, the balanced steps are

\[ \ce{NO2NH2 + H2O <=> NO2NH- + H3O+}, \]

\[ \ce{NO2NH- -> N2O + OH-}, \]

\[ \ce{H3O+ + OH- -> 2H2O}. \]

Here the first equilibrium has been corrected to contain \(\ce{NO2NH-}\).

The rapid acid–base equilibrium gives, at constant water concentration,

\[ [\ce{NO2NH-}] \propto\frac{[\ce{NO2NH2}]}{[\ce{H3O+}]}. \]

Since decomposition of this anion is the rate-limiting step,

\[ v=k_5[\ce{NO2NH-}] \propto\frac{[\ce{NO2NH2}]}{[\ce{H3O+}]}. \]

Thus Mechanism 3 reproduces the experimental rate law. Adding its three steps and cancelling intermediates also gives the correct overall reaction:

\[ \ce{NO2NH2 -> N2O + H2O}. \]

5.3 — Relationship between the rate constants

Applying the rapid-pre-equilibrium approximation to the first step of Mechanism 3,

\[ k_4[\ce{NO2NH2}][\ce{H2O}] =k_{-4}[\ce{NO2NH-}][\ce{H3O+}]. \]

Therefore,

\[ [\ce{NO2NH-}] =\frac{k_4[\ce{H2O}]}{k_{-4}} \frac{[\ce{NO2NH2}]}{[\ce{H3O+}]}. \]

Substitution into the rate expression for the slow step gives

\[ v=k_5[\ce{NO2NH-}] =\frac{k_4k_5[\ce{H2O}]}{k_{-4}} \frac{[\ce{NO2NH2}]}{[\ce{H3O+}]}. \]

Comparison with the experimental law yields

\[ \boxed{k=\frac{k_4k_5[\ce{H2O}]}{k_{-4}}}. \]

Water is the solvent, so its concentration is effectively constant. If it is already included in the forward coefficient, defined by

\[ \widetilde{k}_4=k_4[\ce{H2O}], \]

the same result is written

\[ \boxed{k=\frac{\widetilde{k}_4k_5}{k_{-4}}}. \]

The rapid neutralization constant \(k_6\) does not enter the rate law.

5.4 — Catalysis by hydroxide ions

At constant temperature, using the concentration form of the water ion product,

\[ K_w=[\ce{H3O+}][\ce{OH-}]. \]

Hence

\[ \frac{1}{[\ce{H3O+}]}=\frac{[\ce{OH-}]}{K_w}, \]

and the experimental law becomes

\[ \boxed{v=\frac{k}{K_w}[\ce{NO2NH2}][\ce{OH-}]}. \]

Increasing hydroxide concentration therefore increases the rate.

The catalytic cycle can be expressed as

\[ \ce{NO2NH2 + OH- <=> NO2NH- + H2O}, \]

followed by

\[ \ce{NO2NH- -> N2O + OH-}. \]

In the first step, hydroxide accepts a proton from the amino nitrogen of nitramide; the electron pair of the broken N–H bond remains on nitrogen. The resulting nitramidate ion decomposes in the slow step, regenerating hydroxide. These proton-transfer and regeneration events are not encoded by the SMILES alone.

Cancelling hydroxide and nitramidate gives

\[ \ce{NO2NH2 -> N2O + H2O}. \]

Hydroxide accelerates the reaction while being regenerated, establishing its catalytic role.

5.5 — Pressure as a function of time

Let \(a(t)\) be the concentration of undecomposed nitramide and \(a_0\) its initial value. In the buffered solution,

\[ -\frac{da}{dt}=k'a, \]

so

\[ a(t)=a_0e^{-k't}. \]

One mole of nitramide produces one mole of nitrous oxide. Using the ideal-gas relation, with equal liquid and gas volumes \(V\),

\[ \frac{p(t)V}{RT}=V\bigl(a_0-a(t)\bigr). \]

At complete decomposition,

\[ \frac{p_\infty V}{RT}=Va_0. \]

Dividing gives

\[ \frac{p(t)}{p_\infty} =1-\frac{a(t)}{a_0} =1-e^{-k't}. \]

Therefore,

\[ \boxed{p(t)=p_\infty\left(1-e^{-k't}\right) =(40000\ \mathrm{Pa})\left(1-e^{-k't}\right)}. \]

The units of \(t\) must correspond to those used for \(k'\).

5.6 — Graphical verification

Rearranging the pressure equation gives

\[ \frac{p_\infty-p}{p_\infty}=e^{-k't}, \]

and hence

\[ \boxed{ y=\ln\!\left(\frac{p_\infty}{p_\infty-p}\right)=k't }. \]

The logarithm has a dimensionless argument. From the measurements:

\(t\) / min \(p_\infty-p\) / Pa \(y=\ln[p_\infty/(p_\infty-p)]\)
0 40000 0.00000
5 33200 0.18633
10 27600 0.37106
15 22800 0.56212
20 19200 0.73397
25 16000 0.91629

On the graph, put time in minutes on the horizontal axis and \(y\) on the vertical axis. The six points lie very close to a straight line rising from the origin to approximately \((25,\ 0.916)\).

An unconstrained least-squares fit gives

\[ y=0.00338+(0.03666\ \mathrm{min^{-1}})t, \qquad R^2=0.9998. \]

The small intercept and nearly linear data support the predicted first-order law to the precision of the measurements.

5.7 — Numerical value of \(k'\)

The slope of the graph is \(k'\). Using the initial and final measurements,

\[ k' =\frac{\ln\!\left(\dfrac{40000}{40000-24000}\right)}{25\ \mathrm{min}} =\frac{\ln(2.5)}{25\ \mathrm{min}} =0.03665\ \mathrm{min^{-1}}. \]

This agrees with the slope obtained from all six measurements. Converting to seconds,

\[ k'=\frac{0.03665}{60}\ \mathrm{s^{-1}} =6.109\times10^{-4}\ \mathrm{s^{-1}}. \]

To sensible significant figures,

\[ \boxed{k'\approx3.7\times10^{-2}\ \mathrm{min^{-1}} \approx6.1\times10^{-4}\ \mathrm{s^{-1}}}. \]

The solution was checked against the official solution.