IChO 2005 · Theory problem 7

Kinetic Behaviour of Ozone

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Ozone (\(\ce{O3}\)) is a form of oxygen. It is a natural component of the stratosphere, where it shields the earth from life-destroying ultraviolet radiation. On absorbing light in this region, ozone is converted to dioxygen molecules.

For the overall reaction of ozone decomposition,

Figure 1

One of the proposed mechanisms is expressed as

Figure 2

where \(k_1\), \(k_{-1}\), and \(k_2\) are the rate constants.

7.1 According to the above mechanism what are the differential rate equations for the formation (or consumption) of \(\ce{O3}\), \(\ce{O2}\), and \(\ce{O}\) at time \(t\), assuming step 2 is irreversible.

7.2 Simplification in obtaining the rate law may be found by making appropriate assumptions. Assuming that the concentration of \(\ce{O}\) atoms reaches equilibrium rapidly, its concentration may be given by the equilibrium constant of the reaction (1). The second step is rate determining. Under this equilibrium approximation, deduce the differential rate equation for the \(\ce{O3}\) depletion as a function of \(\ce{O2}\) and \(\ce{O3}\) concentrations.

7.3 Another assumption frequently made is that the rates of oxygen atom production and consumption are equal (this is called steady state). Under the steady state approximation, that is \(d[\ce{O}]/dt=0\), show that the rate equation is: \[ -\frac{d[\ce{O3}]}{dt} = \frac{2k_1k_2[\ce{O3}]^2} {k_{-1}[\ce{O2}]+k_2[\ce{O3}]}. \]

One pathway for the destruction of ozone (\(\ce{2O3 -> 3O2}\)) in the upper atmosphere is catalyzed by Freons. For instance, when \(\ce{CCl2F2}\) (Freon-12) migrates to the upper atmosphere, the ultraviolet photolysis of \(\ce{CCl2F2}\) may give rise to \(\ce{Cl}\) atoms according to the following reaction:

Figure 3

7.4 Chlorine atom can act as a catalyst for the destruction of ozone. The first slow step of a \(\ce{Cl}\)-catalyzed mechanism is proposed as follows:

Figure 4

Assuming a two-step mechanism, propose the second step in the mechanism.

7.5 The activation energy for \(\ce{Cl}\)-catalyzed destruction of ozone is \(2.1\ \mathrm{kJ\,mol^{-1}}\), while the activation energy for the reaction without the presence of catalyst is \(14.0\ \mathrm{kJ\,mol^{-1}}\). Estimate the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25\,^\circ\mathrm{C}\). Assume the frequency factor is the same for each reaction.

Official English text of the problem.

Answers

  • 7.1 \[ \boxed{ \begin{aligned} \frac{d[\ce{O3}]}{dt} &=-k_1[\ce{O3}]+k_{-1}[\ce{O}][\ce{O2}] -k_2[\ce{O3}][\ce{O}],\\ \frac{d[\ce{O2}]}{dt} &=k_1[\ce{O3}]-k_{-1}[\ce{O}][\ce{O2}] +2k_2[\ce{O3}][\ce{O}],\\ \frac{d[\ce{O}]}{dt} &=k_1[\ce{O3}]-k_{-1}[\ce{O}][\ce{O2}] -k_2[\ce{O3}][\ce{O}]. \end{aligned}} \]

  • 7.2 With literal cancellation of the forward and reverse equilibrium rates, \[ \boxed{ -\frac{d[\ce{O3}]}{dt} =\frac{k_1k_2}{k_{-1}} \frac{[\ce{O3}]^2}{[\ce{O2}]} }. \]

  • 7.3 \[ \boxed{ -\frac{d[\ce{O3}]}{dt} =\frac{2k_1k_2[\ce{O3}]^2} {k_{-1}[\ce{O2}]+k_2[\ce{O3}]} }. \]

  • 7.4 \[ \boxed{\ce{ClO(g) + O3(g) -> Cl(g) + 2O2(g)}}. \]

  • 7.5 \[ \boxed{\frac{k_{\mathrm{cat}}}{k_{\mathrm{uncat}}} \approx 1.2\times10^2} \] —a dimensionless ratio of approximately \(120\).

Solution

The species appearing in the mechanism are identified below.

Formula Name SMILES
\(\ce{O3}\) Ozone O=[O+][O-]
\(\ce{O2}\) Dioxygen O=O
\(\ce{O}\) Oxygen atom [O]
\(\ce{Cl}\) Chlorine atom [Cl]
\(\ce{ClO}\) Chlorine monoxide radical Cl[O]
\(\ce{CCl2F2}\) Dichlorodifluoromethane (Freon-12) FC(F)(Cl)Cl
\(\ce{CF2Cl}\) Chlorodifluoromethyl radical F[C](F)Cl

The ozone SMILES depicts one resonance contributor; the actual molecule is bent with equivalent oxygen–oxygen bonds. These SMILES do not specify electronic spin states, including the triplet ground state of dioxygen. The photolysis shown in reaction (3) involves homolytic carbon–chlorine bond cleavage, with one bonding electron going to each radical fragment.

7.1

For the elementary reactions \[ \ce{O3 <=>[k_1][k_{-1}] O + O2}, \qquad \ce{O3 + O ->[k_2] 2O2}, \] define the event rates \[ v_1=k_1[\ce{O3}],\qquad v_{-1}=k_{-1}[\ce{O}][\ce{O2}],\qquad v_2=k_2[\ce{O3}][\ce{O}]. \]

Each forward event of step (1) consumes one ozone molecule and produces one oxygen atom and one dioxygen molecule. The reverse event has the opposite effects. Each event of step (2) consumes one ozone molecule and one oxygen atom and produces two dioxygen molecules. Therefore, \[ \begin{aligned} \frac{d[\ce{O3}]}{dt}&=-v_1+v_{-1}-v_2,\\ \frac{d[\ce{O2}]}{dt}&=v_1-v_{-1}+2v_2,\\ \frac{d[\ce{O}]}{dt}&=v_1-v_{-1}-v_2. \end{aligned} \] Substitution gives the three expanded equations in the answers.

As an atom-balance check, \[ \frac{d}{dt}\left(3[\ce{O3}]+2[\ce{O2}]+[\ce{O}]\right)=0. \]

If concentrations are expressed in \(\mathrm{mol\,L^{-1}}\), \(k_1\) has units \(\mathrm{s^{-1}}\), while \(k_{-1}\) and \(k_2\) have units \(\mathrm{L\,mol^{-1}\,s^{-1}}\).

7.2

For equilibrium in step (1), \[ k_1[\ce{O3}]=k_{-1}[\ce{O}][\ce{O2}], \] so \[ K_c=\frac{[\ce{O}][\ce{O2}]}{[\ce{O3}]} =\frac{k_1}{k_{-1}}, \qquad [\ce{O}]_{\mathrm{eq}} =\frac{k_1[\ce{O3}]}{k_{-1}[\ce{O2}]}. \]

Applying the literal equilibrium cancellation \(v_1=v_{-1}\) to the ozone balance in 7.1 gives \[ -\frac{d[\ce{O3}]}{dt} =v_2=k_2[\ce{O3}][\ce{O}]_{\mathrm{eq}}. \] Hence, \[ \boxed{ -\frac{d[\ce{O3}]}{dt} =\frac{k_1k_2[\ce{O3}]^2}{k_{-1}[\ce{O2}]} }. \]

This expression is second order in ozone and inverse first order in dioxygen. The equilibrium cancellation used here differs from the steady-state condition imposed next.

7.3

The steady-state condition for oxygen atoms is \[ 0=\frac{d[\ce{O}]}{dt} =k_1[\ce{O3}] -\left(k_{-1}[\ce{O2}]+k_2[\ce{O3}]\right)[\ce{O}]. \] Consequently, \[ [\ce{O}]_{\mathrm{ss}} =\frac{k_1[\ce{O3}]} {k_{-1}[\ce{O2}]+k_2[\ce{O3}]}. \]

The same condition can be written \[ v_1-v_{-1}=v_2. \] Substituting this into the ozone balance, \[ -\frac{d[\ce{O3}]}{dt} =v_1-v_{-1}+v_2 =2v_2. \] Thus, \[ \boxed{ -\frac{d[\ce{O3}]}{dt} =2k_2[\ce{O3}][\ce{O}]_{\mathrm{ss}} =\frac{2k_1k_2[\ce{O3}]^2} {k_{-1}[\ce{O2}]+k_2[\ce{O3}]} }. \]

The factor of two has a stoichiometric origin: step (2) consumes one ozone molecule, while net forward reaction through step (1) consumes another ozone molecule to replace the oxygen atom used in step (2).

7.4

The second step must consume the chlorine monoxide intermediate, regenerate the chlorine catalyst, and complete the overall reaction \(\ce{2O3 -> 3O2}\).

Subtracting the given first step, \[ \ce{Cl(g) + O3(g) -> ClO(g) + O2(g)}, \] from the required overall reaction gives \[ \boxed{\ce{ClO(g) + O3(g) -> Cl(g) + 2O2(g)}}. \]

Adding these two steps cancels both \(\ce{Cl}\) and \(\ce{ClO}\), leaving \[ \ce{2O3(g) -> 3O2(g)}. \] The proposed step conserves one chlorine atom and four oxygen atoms. Chlorine is regenerated and is therefore the catalyst; chlorine monoxide is formed and subsequently consumed and is therefore an intermediate.

7.5

The Arrhenius equation is \[ k=A\exp\left(-\frac{E_a}{RT}\right). \] Because the frequency factors are assumed equal, \[ \frac{k_{\mathrm{cat}}}{k_{\mathrm{uncat}}} = \exp\left( \frac{E_{a,\mathrm{uncat}}-E_{a,\mathrm{cat}}}{RT} \right). \]

The activation-energy difference and temperature are \[ \Delta E_a=(14.0-2.1)\ \mathrm{kJ\,mol^{-1}} =11900\ \mathrm{J\,mol^{-1}}, \] \[ T=25+273.15=298.15\ \mathrm{K}. \]

Using \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\), \[ \frac{k_{\mathrm{cat}}}{k_{\mathrm{uncat}}} = \exp\left[ \frac{11900\ \mathrm{J\,mol^{-1}}} {(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})(298.15\ \mathrm{K})} \right] =e^{4.8007} =121.6. \] Therefore, \[ \boxed{\frac{k_{\mathrm{cat}}}{k_{\mathrm{uncat}}} \approx1.2\times10^2}. \]

The solution was checked against the official solution.