Question

Hydrogen bromide is a highly reactive and corrosive gas used mainly as a catalyst for organic reactions. It is produced by reacting hydrogen and bromine gases together. $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) $$ The rate is followed by measuring the intensity of the orange color of the bromine gas. The following data are obtained: $$ \begin{array}{cccc} \hline \text { Expt. } & {\left[\mathrm{H}_{2}\right]} & {\left[\mathrm{Br}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 1 & 0.100 & 0.100 & 4.74 \times 10^{-3} \\ 2 & 0.100 & 0.200 & 6.71 \times 10^{-3} \\ 3 & 0.250 & 0.200 & 1.68 \times 10^{-2} \\ \hline \end{array} $$ (a) What is the order of the reaction with respect to hydrogen, bromine, and overall? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. What are the units for \(k\) ? (d) When \(\left[\mathrm{H}_{2}\right]=0.455 \mathrm{M}\) and \(\left[\mathrm{Br}_{2}\right]=0.215 \mathrm{M},\) what is the rate of the reaction?

Short answer

Based on the given experimental data, the reaction order is determined to be 1 with respect to both hydrogen and bromine, with an overall reaction order of 2. The rate expression for the reaction is given by: $$\text{Rate} = k[\text{H}_2][\text{Br}_2]$$ The rate constant, k, calculated from the experimental data is: $$k = 4.74 \times 10^{-1} \frac{\text{mol}^{-1}\text{L} \cdot \text{s}}$$ For given concentrations of hydrogen and bromine (\([\text{H}_2]=0.455 \mathrm{M}\) and \([\text{Br}_2]=0.215 \mathrm{M}\)), the rate of the reaction is: $$\text{Rate} = 0.046 \frac{\text{mol}}{\text{L} \cdot \text{s}}$$

Step-by-step solution

Determine the order of the reaction with respect to hydrogen and bromine

To determine the order of the reaction, we will compare how the initial rate changes when the concentration of one reactant is changed while keeping the concentration of the other reactant constant. Let's first compare experiments 1 and 2, where the concentration of hydrogen is kept constant: $$\frac{\text{rate}_2}{\text{rate}_1} = \frac{6.71\times10^{-3}}{4.74\times10^{-3}} \approx 1.42$$ The ratio of bromine concentrations in experiment 2 to experiment 1 is: $$\frac{[\text{Br}_2]_2}{[\text{Br}_2]_1} = \frac{0.200}{0.100} \implies 2$$ Since the ratio of the initial rates is not equal to the ratio of the bromine concentrations, the reaction order with respect to bromine is 1. Now let's compare experiments 2 and 3, where the concentration of bromine is kept constant: $$\frac{\text{rate}_3}{\text{rate}_2} = \frac{1.68\times10^{-2}}{6.71\times10^{-3}} \approx 2.5$$ The ratio of hydrogen concentrations in experiment 3 to experiment 2 is: $$\frac{[\text{H}_2]_3}{[\text{H}_2]_2} = \frac{0.250}{0.100} \implies 2.5$$ Since the ratio of the initial rates is equal to the ratio of the hydrogen concentrations, the reaction order with respect to hydrogen is 1. The overall order of the reaction is the sum of the orders with respect to each reactant: \(1+1=2\).

Write the rate expression for the reaction

Now that we know the order of the reaction with respect to each reactant, we can write the rate expression. $$\text{Rate} = k[\text{H}_2]^1[\text{Br}_2]^1$$

Calculate the rate constant, k, and its units

Using the data from any of the experiments, we can calculate the rate constant, k. Let's use data from experiment 1: $$k = \frac{\text{Rate}}{[\text{H}_2][\text{Br}_2]} = \frac{4.74 \times 10^{-3}}{0.100 \times 0.100} = 4.74 \times 10^{-1} \frac{\text{mol}^{-1}\text{L} \cdot \text{s}}$$ The units for k are \(\mathrm{mol}^{-1} \mathrm{L} \cdot \mathrm{s}\).

Calculate the rate of the reaction for given concentrations of H\(_2\) and Br\(_2\)

Now that we have the rate constant, we can calculate the rate of the reaction when \([\text{H}_2]=0.455 \mathrm{M}\) and \([\text{Br}_2]=0.215 \mathrm{M}\): $$\text{Rate} = k[\text{H}_2][\text{Br}_2] = (4.74\times10^{-1} \frac{\text{mol}^{-1}\text{L} \cdot \text{s}})(0.455 \text{M})(0.215 \text{M}) = 0.046 \frac{\text{mol}}{\text{L} \cdot \text{s}}$$ The rate of the reaction under these conditions is 0.046 \(\mathrm{mol} \ \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\).

Key concepts

Rate Law Expression

Understanding the rate law expression is critical when studying the dynamics of a chemical reaction. The rate law indicates how the rate of a reaction is influenced by the concentration of the reactants.
For the reaction between hydrogen and bromine to form hydrogen bromide, we determine the reaction order by analyzing experimental data showing how the reaction rate changes with varying reactant concentrations. The rate law expression has the general form: \[\text{Rate} = k[\text{Reactant}_1]^{n}[\text{Reactant}_2]^{m}\], where \(k\) is the rate constant and \(n\) and \(m\) are the orders with respect to each reactant. The exponents \(n\) and \(m\) are determined experimentally.
In our case, the rate of reaction is first-order with respect to both hydrogen and bromine as their concentration changes lead to proportional changes in the reaction rate. Hence, the rate law is given by: \[\text{Rate} = k[\text{H}_2]^1[\text{Br}_2]^1\].
It’s important for students to grasp that the rate law can only be determined via experiments and not merely from the stoichiometry of the balanced equation. This distinction is essential in predicting how changes in concentration affect the reaction rate in real-time.

Rate Constant Calculation

The rate constant, denoted by \(k\), is a proportionality constant that relates the rate at which a chemical reaction proceeds to the concentration of the reactants. To calculate the rate constant, we use the rate law expression alongside specific experimental data points.
For our reaction, where the rate law is expressed as \[\text{Rate} = k[\text{H}_2][\text{Br}_2]\], we rearrange this to solve for \(k\) based on known rates and reactant concentrations from an experiment: \[k = \frac{\text{Rate}}{[\text{H}_2][\text{Br}_2]}\].
Using data from experiment 1, we find that \[k = 4.74 \times 10^{-1} \mathrm{mol}^{-1} \mathrm{L} \cdot \mathrm{s}\].
The units of the rate constant \(k\), in this case, \(\mathrm{mol}^{-1}\mathrm{L}\cdot\mathrm{s}\), reflect the overall reaction order of 2 (second-order). It is critical to match the units of \(k\) with the reaction order to ensure consistency in calculations. This calculated rate constant is valid only at the specific temperature at which the experiment was conducted, as \(k\) is temperature dependent.

Reactant Concentration Effect

The concentration of reactants in a chemical reaction has a direct effect on the rate of the reaction. According to the rate law, the reaction rate increases as the concentration of the reactants increases. This relationship is particularly important in understanding how a reaction will proceed under different conditions.
For the hydrogen bromide reaction, our rate law expression \[\text{Rate} = k[\text{H}_2]^1[\text{Br}_2]^1\] illustrates that the rate is directly proportional to the concentration of each reactant. When either \([\text{H}_2]\) or \([\text{Br}_2]\) concentrations are doubled, holding the other constant, the initial rate also approximately doubles, reflecting a first-order dependence on each reactant.
This effect is seen when comparing the experimental data. For instance, when \([\text{Br}_2]\) concentration is increased from 0.100 M to 0.200 M, the rate increases from \(4.74 \times 10^{-3}\) mol/L·s to \(6.71 \times 10^{-3}\) mol/L·s. Similarly, altering \([\text{H}_2]\) impacts the rate in a directly proportional way. This proportionality aids in predicting how the reaction will respond to varying concentrations, allowing chemists to control the reaction rate by adjusting reactant levels.