Question

The following data was collected for the destruction of \(\mathrm{O}_{3}\) by \(\mathrm{H}\left(\mathrm{O}_{3}+\mathrm{H} \rightarrow \mathrm{O}_{2}+\mathrm{OH}\right)\) at very low concentrations: $$\begin{array}{llll} \text { Experiment } & {\left[\mathrm{O}_{3}\right], \boldsymbol{M}} & {[\mathrm{H}], M} & \text { Initial Rate, } \boldsymbol{M} / \mathrm{s} \\ \hline 1 & 5.17 \times 10^{-33} & 3.22 \times 10^{-26} & 1.88 \times 10^{-14} \\\ 2 & 2.59 \times 10^{-33} & 3.25 \times 10^{-26} & 9.44 \times 10^{-15} \\ 3 & 5.19 \times 10^{-33} & 6.46 \times 10^{-26} & 3.77 \times 10^{-14} \end{array}$$ (a) Write the rate law for the reaction. (b) Calculate the rate constant.

Short answer

(a) The rate law for the reaction is: $$ \text{rate} = k[\mathrm{O}_3][\mathrm{H}] $$ (b) The rate constant k is: $$ k = 1.18 \times 10^{45} \, \mathrm{M}^{-1}\mathrm{s}^{-1} $$

Step-by-step solution

Determine the order of reaction with respect to O3 (x)

To find the value of x, we can compare experiments 1 and 2 since H is almost constant. Use the formula: $$ \frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[\mathrm{O}_3]_2}{[\mathrm{O}_3]_1}\right)^x $$ Plug in the values from the table: $$ \frac{9.44 \times 10^{-15}}{1.88 \times 10^{-14}} = \left(\frac{2.59 \times 10^{-33}}{5.17 \times 10^{-33}}\right)^x $$

Calculate the value of x

Simplify the equation from step 1 and solve for x: $$ 0.5 = \left(\frac{1}{2}\right)^x $$ Comparing both sides, we can conclude that x = 1.

Determine the order of reaction with respect to H (y)

To find the value of y, we can compare experiments 1 and 3 since O3 is almost constant. Use the formula: $$ \frac{\text{rate}_3}{\text{rate}_1} = \left(\frac{[\mathrm{H}]_3}{[\mathrm{H}]_1}\right)^y $$ Plug in the values from the table: $$ \frac{3.77 \times 10^{-14}}{1.88 \times 10^{-14}} = \left(\frac{6.46 \times 10^{-26}}{3.22 \times 10^{-26}}\right)^y $$

Calculate the value of y

Simplify the equation from step 3 and solve for y: $$ 2 = 2^y $$ Comparing both sides, we can conclude that y = 1.

Write the rate law for the reaction

Now we have x = 1 and y = 1, so the rate law for the reaction is: $$ \text{rate} = k[\mathrm{O}_3][\mathrm{H}] $$

Calculate the rate constant (k)

We can use any of the experiments and the determined rate law to calculate k. Let's use experiment 1: $$ 1.88 \times 10^{-14} = k(5.17 \times 10^{-33})(3.22 \times 10^{-26}) $$ Solve for k: $$ k = \frac{1.88 \times 10^{-14}}{(5.17 \times 10^{-33})(3.22 \times 10^{-26})} = 1.18 \times 10^{45} $$ Solution: (a) The rate law for the reaction is: $$ \text{rate} = k[\mathrm{O}_3][\mathrm{H}] $$ (b) The rate constant k is: $$ k = 1.18 \times 10^{45} \, \mathrm{M}^{-1}\mathrm{s}^{-1} $$

Key concepts

Reaction Order

In chemical kinetics, understanding the order of a reaction is essential to predict how the concentration of reactants affects the rate of reaction. The reaction order refers to the power to which the concentration of a reactant is raised in the rate law equation. In simple terms: it's how changes in reactant concentration influence the rate.
For the decomposition of ozone (\(\mathrm{O}_3\)) in the presence of hydrogen (\(\mathrm{H}\)), we determined the reaction order separately for each reactant. By examining experimental data where the concentration of one reactant remains fairly constant, we can deduce these orders.

• With respect to \(\mathrm{O}_3\), the reaction is first order, as changing its concentration changes the reaction rate proportionally.
• Similarly, with respect to \(\mathrm{H}\), the reaction is also first order.

This means for both reactants, doubling the concentration would double the reaction rate.

Rate Law

The rate law is an equation that shows the relationship between the reactant concentrations and the reaction rate. It is written as:\[\text{rate} = k[\mathrm{O}_3]^x[\mathrm{H}]^y\]where:

• \(k\) is the rate constant
• \([\mathrm{O}_3]\) and \([\mathrm{H}]\) are the concentrations of the reactants \(\mathrm{O}_3\) and \(\mathrm{H}\)
• \(x\) and \(y\) are the reaction orders

As determined in the exercise, both \(x\) and \(y\) have the value of 1 in this particular reaction, making it:\[\text{rate} = k[\mathrm{O}_3][\mathrm{H}]\]This means the reaction rate is dependent on the first power of both \(\mathrm{O}_3\) and \(\mathrm{H}\), indicating a direct relationship with their individual concentrations.

Rate Constant

The rate constant (\(k\)) is a crucial component in the rate law equation that links the reaction rate to the concentration of reactants. It is dependent on the temperature and the specific reaction but remains constant under constant conditions. Essentially, it tells us how fast the reaction will proceed at a given concentration of reactants.
In the provided exercise, the rate constant \(k\) was calculated using the known concentrations of \(\mathrm{O}_3\) and \(\mathrm{H}\), as well as the experimentally determined rate. The calculation was:\[k = \frac{1.88 \times 10^{-14}}{(5.17 \times 10^{-33})(3.22 \times 10^{-26})} = 1.18 \times 10^{45} \, \mathrm{M}^{-1}\mathrm{s}^{-1}\]The units of \(k\) depend on the overall reaction order. For a second-order reaction like this, \(k\) has units of \(\mathrm{M}^{-1}\mathrm{s}^{-1}\). These units ensure that when combined with concentration values, the calculated rate has units of \(\mathrm{M}\) per second.

Experimental Data Analysis

Analyzing experimental data is critical for understanding the kinetics of a reaction. It allows us to determine the rate law and rate constant by examining how varying concentrations influence the reaction rate.
To deduce such relationships, one common approach is to keep the concentration of one reactant constant while varying the other and observe the changes in the reaction rate. By dividing the rates from such experiments, we can eliminate constants and solve for the order of reaction with respect to each reactant.

• For \(\mathrm{O}_3\) and maintaining \(\mathrm{H}\) almost constant, changes in rates revealed a first-order dependency on \(\mathrm{O}_3\).
• Similarly, for \(\mathrm{H}\) with \(\mathrm{O}_3\) constant, the reaction was also first-order in \(\mathrm{H}\).

These observations allow chemists to construct detailed rate laws that model the behavior of the reaction under different conditions, providing a deeper insight into the reaction mechanism and dynamics.