IChO 2026 · Theory problem 2

Kinetics of the Belousov-Zhabotinsky reaction

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T2. Kinetics of the Belousov-Zhabotinsky reaction

6%

2.1 2.2 2.3 2.4 2.5 2.6 2.7 \(\sum\)
2 11 4 2 9 4 3 35

When the chemicals \(\ce{KBrO3}\), \(\ce{CH2(COOH)2}\) (malonic acid, – MA), \(\ce{Ce(SO4)2}\), and \(\ce{H2SO4}\) are mixed, the concentration, \(C\), of Ce species and the colour of the solution oscillates between yellow (due to \(\ce{Ce^4+}\)) and colourless (due to \(\ce{Ce^3+}\)). A schematic picture is given below:

Figure 1

2.1 — 2.0 pt

Write the overall balanced reaction equation of the Belousov-Zhabotinsky reaction (BZ reaction). Note: in this reaction, MA is oxidised to \(\ce{CO2}\), \(\ce{BrO3-}\) is reduced to \(\ce{Br-}\), and Ce(IV) is a catalyst.

The mechanism of the BZ reaction is shown below (BMA – \(\ce{CHBr(COOH)2}\)):

Figure 2
\(k_1=1.0\times10^4\ \mathrm{M^{-2}\,s^{-1}}\)
\(k_2=6.2\times10^4\ \mathrm{M^{-2}\,s^{-1}}\)
\(k_3=4.0\times10^7\ \mathrm{M^{-1}\,s^{-1}}\)
\(k_4=2.0\times10^9\ \mathrm{M^{-2}\,s^{-1}}\)
\(k_5=2.1\ \mathrm{M^{-3}\,s^{-1}}\)
\(k_6=8.2\ \mathrm{M^{-1}\,s^{-1}}\)
\(k_7=1.0\times10^2\ \mathrm{M^{-1}\,s^{-1}}\)

In the BZ reaction, Process C occurs continuously, while Processes A and B alternate with each other.

Assume that:

  • when Process A occurs the solution is yellow and Process B practically does not occur;
  • when Process B occurs the solution is colourless and Process A practically does not occur;
  • \([\ce{BrO3-}]_0=0.06\ \mathrm{M}\), \([\mathrm{MA}]_0=0.1\ \mathrm{M}\), \([\ce{H+}]_0=0.8\ \mathrm{M}\), \([\ce{Ce^4+}]_0=0.001\ \mathrm{M}\);
  • the concentrations of the reactants (except \(\ce{Ce^4+}\)) and pH are maintained constant throughout the BZ reaction.

2.2 — 11.0 pt

Using the steady-state approximation, calculate the stationary molar concentrations of \(\ce{HBrO2}\) in Processes A and B, \([\ce{HBrO2}]_{\mathrm A}\) and \([\ce{HBrO2}]_{\mathrm B}\).

If you could not calculate \([\ce{HBrO2}]_{\mathrm A}\) and \([\ce{HBrO2}]_{\mathrm B}\), you can use the values \([\ce{HBrO2}]_{\mathrm A}=1\times10^{-5}\ \mathrm{M}\) and \([\ce{HBrO2}]_{\mathrm B}=1\times10^{-10}\ \mathrm{M}\) for the following calculations.

During the BZ reaction, the concentrations of \(\ce{HBrO2}\) and \(\ce{Br-}\) oscillate strictly in relation to each other, and this is sketched in the phase portrait of \([\ce{HBrO2}]\) vs \([\ce{Br-}]\):

Figure 3

Here \([\ce{Br-}]_{\max}\) is the maximum concentration and \([\ce{Br-}]_{\mathrm{critical}}\) is the critical concentration of \(\ce{Br-}\) at which there is a switch over in processes from A to B or from B to A. To switch Process A to Process B, the reaction rate of the elementary step (4) must exceed that of the elementary step (1) and vice versa.

2.3 — 4.0 pt

Calculate \([\ce{Br-}]_{\mathrm{critical}}\).

If you could not calculate \([\ce{Br-}]_{\mathrm{critical}}\), use \([\ce{Br-}]_{\mathrm{critical}}=1\times10^{-7}\ \mathrm{M}\) for the following calculations.

2.4 — 2.0 pt

In which direction in the phase portrait of \([\ce{HBrO2}]\) vs \([\ce{Br-}]\) will the concentrations of \([\ce{HBrO2}]\) and \([\ce{Br-}]\) change over time? Tick the correct box.

Figure 4

The system does not traverse the phase portrait of \([\ce{HBrO2}]\) vs \([\ce{Br-}]\) at a constant velocity, that is the concentration of \([\ce{Br-}]\) slowly decreases from \([\ce{Br-}]_{\max}=7.0\times10^{-4}\ \mathrm{M}\) to \([\ce{Br-}]_{\mathrm{critical}}\), and then almost immediately reaches \([\ce{Br-}]_{\max}\) again.

2.5 — 9.0 pt

Calculate based on the data the period of oscillations, \(\tau\), of the BZ reaction, in seconds.

2.6 — 4.0 pt

What will be the effect if the following actions (1-4) are performed on an open system where a BZ reaction is taking place? Tick only one box (a-e) for each case.

1. Adding of small amount of \(\ce{Ce^4+}\) during Process A.
2. Adding of a small amount of \(\ce{Ag+}\) during Process B.
3. Adding of a small amount of \(\ce{Br-}\) during Process B.
4. Continuous addition of \(\ce{Br-}\).		 [ a ] prolongs Process A
[ b ] prolongs Process B
[ c ] Process A switches to Process B
[ d ] Process B switches to Process A
[ e ] oscillations stop

2.7 — 3.0 pt

Which graph qualitatively illustrates the rate of change of Gibbs energy \(\left(\frac{dG}{dt}\right)\) of a closed system in which an oscillatory reaction occurs? Tick the correct box.

Figure 5

Official English text of the problem.

Answers

  • 2.1: \(\ce{3CH2(COOH)2 + 4BrO3- -> 9CO2 + 4Br- + 6H2O}\).
  • 2.2: \([\ce{HBrO2}]_{\mathrm A}=\boxed{6.0\times10^{-6}\ \mathrm{M}}\); \([\ce{HBrO2}]_{\mathrm B}=\boxed{5.0\times10^{-11}\ \mathrm{M}}\).
  • 2.3: \([\ce{Br-}]_{\mathrm{critical}}=\boxed{3.0\times10^{-7}\ \mathrm{M}}\).
  • 2.4: Upper-right diagram of the answer-sheet figure for 2.4: clockwise traversal.
  • 2.5: \(\boxed{\tau\approx48\ \mathrm{s}}\).
  • 2.6: Intended choices: 1–c; 2–d; 3–b; 4–e, subject to sufficient perturbation strength as explained below.
  • 2.7: Bottom-right graph of the answer-sheet figure for 2.7: \(dG/dt\) undergoes damped oscillations below zero and approaches zero.

Solution

The species used below have the following names and SMILES.

Species Name SMILES
\(\ce{CH2(COOH)2}\), MA Malonic acid O=C(O)CC(=O)O
\(\ce{CHBr(COOH)2}\), BMA Bromomalonic acid O=C(O)C(Br)C(=O)O
\(\ce{BrO3-}\) Bromate ion [O-][Br](=O)=O
\(\ce{Br-}\) Bromide ion [Br-]
\(\ce{HBrO2}\) Bromous acid O[Br]=O
\(\ce{HBrO}\) Hypobromous acid OBr
\(\ce{BrO2^{\bullet}}\) Bromine dioxide radical O=[Br]=O
\(\ce{CO2}\) Carbon dioxide O=C=O
\(\ce{H2O}\) Water O
\(\ce{H+}\) Hydrogen ion (proton) [H+]
\(\ce{Ce^4+}\) Cerium(IV) ion [Ce+4]
\(\ce{Ce^3+}\) Cerium(III) ion [Ce+3]
\(\ce{Ag+}\) Silver(I) ion [Ag+]
\(\ce{AgBr}\) Silver bromide [Ag+].[Br-]

The silver bromide SMILES specifies its constituent ions, not the solid’s crystal lattice. The ionic SMILES do not show solvation.

2.1

Complete oxidation of one molecule of malonic acid to three molecules of carbon dioxide releases eight electrons:

\[ \ce{CH2(COOH)2 + 2H2O -> 3CO2 + 8H+ + 8e-}. \]

Reduction of bromate to bromide consumes six electrons:

\[ \ce{BrO3- + 6H+ + 6e- -> Br- + 3H2O}. \]

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4. Cancelling 24 electrons, 24 hydrogen ions, and the common water gives

\[ \boxed{\ce{3CH2(COOH)2 + 4BrO3- -> 9CO2 + 4Br- + 6H2O}}. \]

Cerium cycles between its oxidation states and is regenerated, so it does not appear in the net equation.

2.2

Define

\[ x=[\ce{HBrO2}],\qquad r=[\ce{BrO2^{\bullet}}],\qquad b=[\ce{Br-}], \] \[ a=[\ce{BrO3-}]=0.06\ \mathrm{M},\qquad h=[\ce{H+}]=0.8\ \mathrm{M}. \]

Let \(v_i\) denote the reaction rate of elementary step \(i\), with stoichiometric factors included separately in species balances.

For Process A,

\[ v_1=k_1xah,\qquad v_2=k_2r[\ce{Ce^3+}]h,\qquad v_3=k_3x^2. \]

Applying the steady-state approximation to the bromine dioxide radical gives

\[ \frac{dr}{dt}=2v_1-v_2\simeq0 \quad\Longrightarrow\quad v_2=2v_1. \]

The bromous-acid balance is

\[ \frac{dx}{dt}=-v_1+v_2-2v_3\simeq0. \]

The factor 2 before \(v_3\) is essential: each occurrence of step (3) consumes two molecules of bromous acid. Substituting \(v_2=2v_1\),

\[ k_1xah-2k_3x^2=0. \]

The nonzero stationary concentration corresponding to active Process A is therefore

\[ x_{\mathrm A}=\frac{k_1ah}{2k_3}. \]

Numerically,

\[ x_{\mathrm A} = \frac{ (1.0\times10^4\ \mathrm{M^{-2}s^{-1}}) (0.06\ \mathrm M)(0.8\ \mathrm M) }{ 2(4.0\times10^7\ \mathrm{M^{-1}s^{-1}}) } = \boxed{6.0\times10^{-6}\ \mathrm M}. \]

For Process B, step (5) produces bromous acid and step (4) consumes it:

\[ v_4=k_4xbh,\qquad v_5=k_5abh^2. \]

Thus

\[ \frac{dx}{dt}=v_5-v_4\simeq0, \]

and, cancelling the nonzero factors \(b\) and \(h\),

\[ x_{\mathrm B}=\frac{k_5ah}{k_4}. \]

Hence

\[ x_{\mathrm B} = \frac{ (2.1\ \mathrm{M^{-3}s^{-1}}) (0.06\ \mathrm M)(0.8\ \mathrm M) }{ 2.0\times10^9\ \mathrm{M^{-2}s^{-1}} } = 5.04\times10^{-11}\ \mathrm M \approx \boxed{5.0\times10^{-11}\ \mathrm M}. \]

2.3

At the switching threshold, steps (1) and (4) have equal rates:

\[ k_1xah=k_4xb_{\mathrm{critical}}h. \]

Cancelling \(x\) and \(h\) gives

\[ b_{\mathrm{critical}} =\frac{k_1a}{k_4} = \frac{1.0\times10^4}{2.0\times10^9} (0.06\ \mathrm M) = \boxed{3.0\times10^{-7}\ \mathrm M}. \]

Below this concentration, step (1) dominates and favours Process A. Above it, step (4) dominates and favours Process B.

2.4

During Process A, the solution contains appreciable cerium(IV), so Process C produces bromide. The bromide concentration increases along the upper branch until the system switches to Process B.

During Process B, bromide is consumed while the bromous-acid concentration remains low. The system moves left along the lower branch until bromide falls to the switching threshold and Process A resumes.

Consequently, the arrows must point:

  • right along the upper horizontal edge;
  • down and right along the right sloping edge;
  • left along the lower horizontal edge;
  • up and left along the left sloping edge.

This is clockwise traversal, shown in the upper-right diagram of the answer-sheet figure for 2.4.

2.5

The period is dominated by the slow decrease of bromide during Process B. The subsequent return to the maximum concentration is taken to require negligible time.

The full bromide balance during B is

\[ \frac{db}{dt}=-v_4-v_5+v_7. \]

In the intended approximation, the colourless B interval has very little cerium(IV), so

\[ v_7=k_7[\ce{Ce^4+}][\mathrm{BMA}] \]

is neglected during this slow interval. From the steady-state calculation, \(v_4=v_5\). Therefore,

\[ \frac{db}{dt}\simeq-2v_5=-2k_5ah^2b. \]

Both steps (4) and (5) consume bromide; this supplies the factor 2.

The effective first-order decay constant is

\[ k_{\mathrm{eff}} =2k_5ah^2 =2(2.1\ \mathrm{M^{-3}s^{-1}}) (0.06\ \mathrm M)(0.8\ \mathrm M)^2 =0.16128\ \mathrm{s^{-1}}. \]

Integration gives

\[ b(t)=b_{\max}e^{-k_{\mathrm{eff}}t}. \]

Setting \(b(\tau)=b_{\mathrm{critical}}\),

\[ \tau \simeq \frac{1}{k_{\mathrm{eff}}} \ln\left(\frac{b_{\max}}{b_{\mathrm{critical}}}\right). \]

Thus

\[ \tau = \frac{ \ln\left(\dfrac{7.0\times10^{-4}}{3.0\times10^{-7}}\right) }{ 0.16128\ \mathrm{s^{-1}} } =48.0844\ \mathrm s \approx \boxed{48\ \mathrm s}. \]

2.6

  1. Adding cerium(IV) during Process A: (c).
    Increasing \([\ce{Ce^4+}]\) accelerates Process C and therefore bromide production. If this raises bromide through its critical concentration, step (4) overtakes step (1), and Process A switches to Process B.

  2. Adding silver(I) during Process B: (d).
    Silver(I) removes bromide by precipitation: \[ \ce{Ag+ + Br- -> AgBr(s)}. \] If bromide falls below its critical concentration, step (1) becomes dominant, and Process B switches to Process A.

  3. Adding bromide during Process B: (b).
    The bromide concentration is moved farther above the switching threshold. More time is needed to consume it down to that threshold, so the addition prolongs Process B.

  4. Continuous addition of bromide: (e).
    A sufficient continuous supply maintains bromide above the critical concentration, sustaining Process B and preventing the alternating switches. The oscillations stop.

2.7

For a spontaneously reacting closed system at constant temperature and pressure,

\[ \frac{dG}{dt}\leq0. \]

The concentrations can oscillate, and the rate of Gibbs-energy decrease can vary, while Gibbs energy itself continues to decrease. As equilibrium is approached,

\[ \frac{dG}{dt}\longrightarrow0. \]

The intended graph therefore shows oscillations in \(dG/dt\) that remain below zero and diminish as the curve approaches zero: the bottom-right graph of the answer-sheet figure for 2.7.

The solution was checked against the official solution.