Question

\(\begin{aligned} \text { Consider the reaction. } \\ & 2 \operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \end{aligned}\) \begin{equation} \begin{array}{l}{\text { a. Express the rate of the reaction in terms of the change in concen- }} \\ {\text { tration of each of the reactants and products. }}\end{array} \end{equation} \begin{array}{l}{\text { b. In the first } 25.0 \text { s of this reaction, the concentration of HBr drops }} \\ {\text { from } 0.600 \mathrm{M} \text { to } 0.512 \mathrm{M} \text { . Calculate the average rate of the reac- }} \\\ {\text { tion during this time interval. }} \\ {\text { c. If the volume of the reaction vessel in part b is } 1.50 \mathrm{L}, \text { what }} \\ {\text { amount of } \mathrm{Br}_{2}(\text { in moles) forms during the first } 15.0 \mathrm{s} \text { of the }} \\ {\text { reaction? }}\end{array}

Short answer

a. Rate = -\(\frac{1}{2}\)\(\frac{\Delta[HBr]}{\Delta t}\) = \(\frac{\Delta[H_2]}{\Delta t}\) = \(\frac{\Delta[Br_2]}{\Delta t}\). b. The average rate of the reaction is -0.00176 M/s. c. The amount of Br2 formed in the first 15 seconds is 0.0198 moles.

Step-by-step solution

Express the Rate of Reaction

To express the rate of the reaction in terms of the concentration change of reactants and products, we use the coefficients from the balanced chemical equation to establish a relationship. The rate of disappearance of HBr is twice the rate of appearance of both H2 and Br2 because for every 2 moles of HBr consumed, 1 mole of H2 and 1 mole of Br2 are formed. Thus \[\text{Rate} = -\frac{1}{2} \frac{\Delta[HBr]}{\Delta t} = \frac{\Delta[H_2]}{\Delta t} = \frac{\Delta[Br_2]}{\Delta t}\].

Calculate the Average Rate of Reaction

Using the concentration change of HBr over the time interval given, the average rate can be calculated by \[\text{Average rate} = -\frac{1}{2} \frac{\Delta[HBr]}{\Delta t}\] \[\text{Average rate} = -\frac{1}{2} \frac{0.600 M - 0.512 M}{25.0 s}\] \[\text{Average rate} = -\frac{1}{2} \frac{0.088 M}{25.0 s}\] \[\text{Average rate} = -\frac{0.044 M}{25.0 s}\] \[\text{Average rate} = -0.00176 M/s\]. The negative sign indicates the concentration of HBr is decreasing.

Calculate the Amount of Br2 Formed

First calculate the change in moles of HBr over 15.0 s, then use the stoichiometry to find moles of Br2. \[\text{Change in [HBr]} = \text{Average rate} \times \text{time}\] \[\text{Change in [HBr]} = -0.00176 M/s \times 15.0 s\] \[(\text{Change in moles of HBr}) \times (1.50 L) = 0.00176 M \times 15.0 s \times 1.50 L\] \[(\text{Change in moles of HBr}) = \frac{0.00176 M \times 15.0 s \times 1.50 L}{1 L}\] \[(\text{Change in moles of HBr}) = 0.0396 moles\] Since the reaction produces 1 mole of Br2 for every 2 moles of HBr, \[(\text{Change in moles of Br2}) = \frac{1}{2} \times (\text{Change in moles of HBr})\] \[(\text{Change in moles of Br2}) = \frac{1}{2} \times 0.0396 moles \] \[(\text{Change in moles of Br2}) = 0.0198 moles\].

Key concepts

Rate of Reaction

Understanding the rate of a chemical reaction is crucial for grasping kinetics, which is the study of how fast chemical processes occur. The rate of reaction is defined as the speed at which reactants turn into products. It's typically expressed in terms of the change in concentration of a reactant or product per unit time, such as moles per liter per second ((M/s)). A faster rate means a quicker change in concentration.

In the given exercise, the rate is tied to the consumption of hydrobromic gas (HBr(g)) and the formation of hydrogen gas (H_2(g)) and bromine gas (Br_2(g)). The balanced chemical equation ensures the rate at which HBr disappears is double the rate at which both H_2 and Br_2 appear. This relationship is expressed using stoichiometry coefficients from the balanced equation.

Concentration Change

The concentration of substances in a reaction mixture is integral to determining the reaction's progress. Change in concentration over time can indicate how quickly a reactant is being used up or how fast a product is being formed. In a closed system where volume remains constant, concentration changes are directly related to the number of moles of substances reacting.

In the exercise, the concentration of HBr decreases, which is a clear indicator of the reaction moving forward. By tracking this change, we can calculate the reaction rate. The concentration change is also vital for understanding the reaction kinetics, as it may help in determining the order of reaction and the rate law.

Reaction Stoichiometry

Reaction stoichiometry deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. These relationships are derived from the balanced chemical equation. Stoichiometry allows predicting how much product will form from a given amount of reactants or how much reactants are needed to produce a certain amount of product.

In this case, the stoichiometry involves the 2:1:1 ratio of HBr to H_2 to Br_2 as prescribed by the balanced equation. This means two moles of HBr are needed to produce one mole each of H_2 and Br_2. Stoichiometry is the tool that allows us to determine, for example, the amount of Br_2 produced when a certain amount of HBr is consumed.

Average Rate Calculation

Calculating the average rate of a reaction involves measuring the change in concentration of a reactant or product over a specific time period. It is a simplistic way to describe the reaction rate, given that real reaction rates can vary at different stages of the process.

The formula for the average rate calculation is \(Average rate = \frac{-1}{ stoichiometric coefficient} \frac{\Delta[Concentration]}{\Delta t}\), where \Delta[Concentration] is the change in concentration and \Delta t is the change in time. In our exercise, we focus on the consumption of HBr, hence the negative sign reflects the decrease in its concentration. This simple formula provides a way to quantify the reaction's progress over the given time interval.