Question

Consider the reaction $$3 \mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E}$$ where the rate law is defined as $$-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}]$$ An experiment is carried out where \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \space M\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[A]=3.26 \times 10^{-5} M,\) calculate the value of \(k\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(\mathrm{B}\) and the concentration of A after 10.0 min.

Short answer

The rate constant (k) for the given reaction is approximately \(3.52 \times 10^{-4} \space M^{-2}s^{-1}\). The half-life of the experiment is about 711.0 seconds. After 10.0 minutes, the concentration of A is approximately \(1.55 \times 10^{-5} M\) and the concentration of B is approximately 0.99997 M.

Step-by-step solution

Find the rate constant (k)

Given the rate law: \[ -\frac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] \] We are also provided with the information that at 3.00 min: \[ [\mathrm{A}] = 3.26 \times 10^{-5} M \] Using this information, we can now find the value of k: \[ \frac{1.00 \times 10^{-4} M - 3.26 \times 10^{-5} M }{180 \space s} = k (3.26 \times 10^{-5} M)^2(1.00 M)(1.00 M) \] Calculating the value for k: \[ k = \frac{6.74 \times 10^{-5} M}{180 \space s \times 1.062 \times 10^{-9} M^3} \approx 3.52 \times 10^{-4} M^{-2}s^{-1} \]

Calculate the half-life

To find the half-life of this reaction, we need to determine the time when the concentration of A is half of its initial concentration: \[ \frac{1}{2} [\mathrm{A}]_0 = \frac{1}{2}(1.00 \times 10^{-4} M) = 5.00 \times 10^{-5} M \] Now, we will use the rate law to find out the time when the concentration of A reaches half of its initial concentration. We will consider the initial concentration of B and C as constant for simplification: \[ -\frac{\Delta [\mathrm{A}]}{\Delta t} = k[\mathrm{A}]^2[\mathrm{B}]_0[\mathrm{C}]_0 \] Rearranging the equation and integrating: \[ \int_{[\mathrm{A}]_0}^{\frac{1}{2}[\mathrm{A}]_0} \frac{d[\mathrm{A}]}{[\mathrm{A}]^2} = \int_0^t -k[\mathrm{B}]_0[\mathrm{C}]_0 dt \] After integrating, we get: \[ -\frac{1}{[\mathrm{A}]} \Big|_{[\mathrm{A}]_0}^{\frac{1}{2}[\mathrm{A}]_0} = -k[\mathrm{B}]_0[\mathrm{C}]_0 t \] Now substituting the initial concentration values and the value of k we found in Step 1, we can calculate the half-life: \[ t_{1/2} = \frac{1}{2(3.52 \times 10^{-4} \space M^{-2}s^{-1})(1.00 M)(1.00 M)} \approx 711.0 s \]

Calculate the concentration of A and B after 10.0 minutes

To find the concentration of A after 10 minutes, we can use the following equation derived from our integrated rate law: \[ \frac{1}{[\mathrm{A}]} = kt[\mathrm{B}]_0[\mathrm{C}]_0 + \frac{1}{[\mathrm{A}]_0} \] Using the provided data and the value of k from Step 1: \[ \frac{1}{[\mathrm{A}]} = (3.52 \times 10^{-4}M^{-2}s^{-1})(600s)(1.00M)(1.00M) + \frac{1}{1.00 \times 10^{-4}M} \] Solving for the concentration of A after 10 minutes: \[ [\mathrm{A}] \approx 1.55 \times 10^{-5} M \] Now we will use stoichiometry to calculate the concentration of B after 10 minutes: \[ \Delta[\mathrm{B}] = -\frac{1}{3} \Delta[\mathrm{A}] = -\frac{1}{3}(1.00 \times 10^{-4} - 1.55 \times 10^{-5}) = -2.82 \times 10^{-5} \] The concentration of B after 10 minutes: \[ [\mathrm{B}] = [\mathrm{B}]_0 + \Delta[\mathrm{B}] = 1.00M - 2.82 \times 10^{-5}M \approx 0.99997 M \]

Key concepts

Chemical Kinetics

Chemical kinetics is the study of the speed or rate at which chemical reactions occur, as well as how they happen. The rate of a chemical reaction is crucial because it tells us how quickly reactants are transformed into products. This has practical implications in various fields, from pharmaceuticals to materials science.

In our exercise, we are dealing with a reaction involving species A, B, and C producing D and E. The rate at which A is consumed is described by the rate law, which shows that it's dependent on the concentrations of A, B, and C, and on a proportionality constant known as the rate constant, denoted as 'k'. This indicates the consumption of A is second-order with respect to its own concentration, and first-order with respect to the concentrations of B and C. Understanding this relationship allows chemists to predict how the reaction speed will change if the concentration of one or more reactants changes.

Rate Constant Calculation

The rate constant, symbolized by 'k', is a parameter that quantifies the speed of a chemical reaction. This constant is essential for predicting the behavior of reactions under various conditions. For our particular chemical reaction, we used given concentrations and the change in concentration over time to find 'k'.

The rate law for this reaction is expressed as \(-\frac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}]\), meaning the rate is proportional to the square of the concentration of A and to the concentrations of B and C. By rearranging and integrating this rate law, chemists can determine 'k' from experimental data, enabling the prediction of how fast the reaction will proceed under different circumstances. In essence, knowing the rate constant helps in calculating how much product can be formed over a given time frame and is fundamental in the design of chemical processes and reactors.

Half-Life of a Reaction

The half-life of a reaction is the time it takes for half of the reactant to be used up or for the concentration to fall to half its initial value. In the context of our reaction, the half-life depends on the concentration of the reactants and the rate constant 'k'. Notably, for reactions where the order with respect to one of the reactants is more than one, the half-life will change over time as the concentration of that reactant changes.

The solution's approach to calculating the half-life focused on the integrated rate law. By setting up an integral from the initial concentration to half that value, we can isolate the half-life. In practical terms, understanding the half-life of a reaction is valuable in various disciplines, including pharmacology, environmental science, and in any industry where controlling the rate of a chemical reaction is necessary. It provides essential insights into the duration over which reactants are effective or how long it takes for a reaction to reach a certain point of completion.