# Solutions 8: Chemical Kinetics

- Page ID
- 10761

**These are solutions to select problems from homework set #8.**

### S8.1

A) This is a little complicated question and requires students to identify that the rate law can involve both reactants and products (since the reaction is reversible they are coupled). Since there are elementary reactions (or steps in a overall reaction mechanism):

\[\text{rate} = k_f[SO_2][CO] − k_r[SO][CO_2] \tag{Rx: 1}\]

\[\text{rate} = k_f[SO][CO] − k_r[S][CO_2] \tag{Rx. 2}\]

\[\text{rate} = k_f[SO][SO_2] − k_r[S][SO_3] \tag{Rx. 3}\]

Where \(k_f\) is the rate constant for the forward reaction and \(k_r\) is the rate constant for the reverse reaction.

B)

\[ K_c=\dfrac{[SO][CO_2]}{[SO_2][CO]} \tag{Rx. 1}\]

\[ K_c=\dfrac{[S][CO_2]}{[SO][CO]} \tag{Rx. 2}\]

\[ K_c=\dfrac{[S][SO_3]}{[SO][SO_2]} \tag{Rx. 3}\]

C) At equilibration, the rate of the forward reaction is equal to the rate of the reverse reaction.

\[ k_f[SO_2][CO] = k_r[SO][CO_2] \tag{Rx. 1}\]

\[ k_f[SO][CO] = k_r[S][CO_2] \tag{Rx. 2}\]

\[k_f[SO][SO_2] = k_r[S][SO_3] \tag{Rx. 3}\]

So

\[ K_c=\dfrac{[SO][CO_2]}{[SO_2][CO]} =\dfrac{k_f}{k_r} \tag{Rx. 1}\]

\[ K_c=\dfrac{[S][CO_2]}{[SO][CO]} =\dfrac{k_f}{k_r} \tag{Rx. 2}\]

\[ K_c=\dfrac{[S][SO_3]}{[SO][SO_2]} =\dfrac{k_f}{k_r} \tag{Rx. 3}\]

### S8.2

no solution

### S8.3

The reaction is second order: first order in O and first order in O_{3}. Ozone is being produced faster than it is being destroyed. If ozone concentrations are not increasing, then either some other reaction must be consuming some of the ozone produced in this reaction or the ozone-producing reaction does not operate at this rate continuously.

##### S8.4

no solution

### S8.5

Yes; the object is about 2300 yr old.

### S8.6

no solution

### S8.7

*k*= 0.05626 yr^{−1}- 0.487 g of
^{3}H

### S8.8

no solution

### S8.9

no solution

### S8.10

no solution

### S8.11

- second order, first order in O and first order in O
_{3}; - 17 kJ/mol;
- 0.44 kJ/mol;

- Cl is a potent catalyst for ozone destruction because there is a large decrease in
*E*_{a}when Cl is present.

### S8.13

\(Ni_{1−x}O\) is a nonstoichiometric oxide that contains a fraction of Ni(I) sites. These can react with oxygen to form a Ni(III)-oxide site, which is reduced by CO to give Ni(I) and CO_{2}.

### S8.17

0.35 g of ^{32}P

### S8.19

4.1 × 10^{15}

### S8.20

A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction.

### S8.22

In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased.

### S8.24

- Heterogeneous catalysts are easier to recover.
- Collision frequency is greater for homogeneous catalysts.
- Homogeneous catalysts are often more sensitive to temperature.
- Homogeneous catalysts are often more expensive.

### S8.26

The \(Mn^{2+}\) ion donates two electrons to \(Ce^{4+}\), one at a time, and then accepts two electrons from \(Tl^+\). Because \(Mn\) can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions.

### S8.29

\(\frac{\Delta[\textrm{ES}]}{\Delta t}=-(k_2+k_{-1})[\textrm{ES}]+k_1[\textrm E][\textrm S]+k_{-2}[\textrm E][\textrm P]\approx 0\)

### S8.31

In both cases, the product of pathway \(A\) is favored. All of the \(Z\) produced in the catalyzed reversible pathway \(B\) will eventually be converted to \(X\) as \(X\) is converted irreversibly to \(Y\) by pathway \(A\).

\[Z\underset{\textrm B}{\rightleftharpoons}X\xrightarrow{\textrm A}Y\]

### S8.33

So, given the overall reaction and the experimentally determined rate law, we must determine which of the proposed mechanisms will yield the same rate law, making it a possible mechanism for the overall reaction.

**Mechanism 1**

This mechanism one is a one-step reaction, which means it is elementary. This makes determining its rate law a piece of chemistry cake, which is the best flavor. For elementary reactions, the reactants in the equation appear in the rate law expression and their coefficients become the exponents. So,

\[1X_2 + 1Z_2 \rightarrow 2XZ\]

\[Rate = k[X_2]^1[Z_2]^1\]

Huzzah! a) is a possible mechanism for the total reaction. But let us not ignore mechanism two.

**Mechanism 2**

For a multi-step mechanism, we first look at the slow step. The slow step is the rate-determining step, which is important for determining the rate law expression. Let us label the steps of mechanism b):

\[Z_2 \overset{k_1}{\underset{k_{-1}}\rightleftharpoons} 2Z \tag{Step 1: fast}\]

\[Z + X_2 \overset{k_2}{\underset{k_{-2}}\rightleftharpoons} ZX_2 \tag{Step 2: fast}\]

\[ZX_2 + Z \overset{k_3}\rightarrow 2XZ \tag{Step 3: slow}\]

Both 1) and 2) are reversible reactions, so they have \(k\) values in both direction. Recall equilibrium constants:

\[K_1=\dfrac{k_1}{k_{-1}}=\dfrac{[Z]^2}{[Z_2]}\]

Knowing the slow step determined the rate law (as the rate determining step):

Rate law for step 3 would then be

\[Rate = k_3[ZX_2][Z]\]

which does not match

\[Rate = k[X_2][Z_2]\]

But let us take a look at steps 1 and 2.

Because both are reversible reactions, they have \(K\) constants, as such:

\[K_1=\dfrac{k_1}{k_{-1}}=\dfrac{[Z]^2}{[Z_2]}\]

\[K_2=\dfrac{k_2}{k_{-2}}=\dfrac{[ZX_2]}{[Z][X_2]}\]

Now we can substitute things in from the above equations into the rate law expression for 3) \(Rate = k_3[ZX_2][Z]\)

\(K_2\):

\(\dfrac{k_2}{k_{-2}}=\dfrac{[ZX_2]}{[Z][X_2]}\)

\([ZX_2]=\dfrac{k_2}{k_{-2}}{[Z][X_2]}\)

So,

\(Rate = k_3[ZX_2][Z] = \dfrac{k_3k_2}{k_{-2}}{[Z][X_2]}[Z]\)

\(Rate = \dfrac{k_3k_2}{k_{-2}}{[Z]^2[X_2]}\)

and now we are getting somewhere.

\(K_1\):

\(K_1=\dfrac{k_1}{k_{-1}}=\dfrac{[Z]^2}{[Z_2]}\)

\([Z]^2=\dfrac{k_1}{k_{-1}}{[Z_2]}\)

So,

\(Rate = \dfrac{k_3k_2}{k_{-2}}{[Z]^2[X_2]}\)

\(Rate = \dfrac{k_3k_2k_1}{k_{-2}k_{-1}}{[Z_2][X_2]}\)

and \(\dfrac{k_3k_2k_1}{k_{-2}k_{-1}}\) is just a combination of \(k\) constants, so we can reduce it down to

\[Rate = k{[X_2][Z_2]}\]

which now matches the experimentally determined rate law expression.

**Upshot**:

Hence both mechanism both are possible mechanisms for the overall reaction given the experimentally determined rate law. More data is needed to identify one as the relevant reaction mechanism (if possible).