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 \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[\mathrm{A}]=3.26 \times 10^{-5} M,\) calculate the value of \(k .\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(B\) and the concentration of A after 10.0 min.

Short answer

The value of rate constant k is \(6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1}\). The half-life for this experiment is 15.4 minutes. After 10.0 minutes, the concentration of A is \(5.16 \times 10^{-5} \mathrm{M}\), and the concentration of B is \(0.999984\, \mathrm{M}\).

Step-by-step solution

Calculate the value of k

We are given the initial concentration of A, B, and C, as well as the final concentration of A after 3 minutes. We can use this information and the rate law to find the value of k. We can set up the rate equation as follows: $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] $$ Plug in the given values: $$ -\frac{(3.26 \times 10^{-5} \text{ M} - 1.00 \times 10^{-4} \text{ M})}{(3.00 \text{ min})} =k\left(1.00 \times 10^{-4} \text{ M}\right)^{2}(1.00 \mathrm{M})(1.00 \mathrm{M}) $$ Now, we can solve for value of k: $$ k=\frac{1}{12}\left(\frac{7.74 \times 10^{-5} \text{ M}}{(1.00 \times 10^{-4})^{2} \text{ M}^2}\right)= 6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1}. $$ So, the value of k is \(6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1}\).

Calculate the half-life of this experiment

We can use the equation for the half-life of a reaction, defined as: $$ t_{1/2} = \frac{1}{k [\mathrm{A}]_0 [\mathrm{B}]_0 [\mathrm{C}]_0} $$ Plug in the values we have found so far: $$ t_{1/2} = \frac{1}{(6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1})(1.00 \times 10^{-4} \mathrm{M})(1.00 \mathrm{M})(1.00 \mathrm{M})} = 15.4 \, \mathrm{min} $$ The half-life for this experiment is 15.4 minutes.

Calculate the concentration of A and B after 10.0 min

To find the concentration of A and B at 10.0 minutes, we can use the integrated rate law for this reaction, which is: $$ [\mathrm{A}]_{t}=\frac{1}{kt+1 /[\mathrm{A}]_{0}} $$ Plug in the values for A: $$ [\mathrm{A}]_{10}=\frac{1}{(6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1})(10 \, \mathrm{min})+1 /(1.00 \times 10^{-4}\, \mathrm{M})}= 5.16 \times 10^{-5}\, \mathrm{M} $$ Now, we can find the concentration of B at 10.0 minutes using stoichiometry. Since we started with 1.00 M of B and the stoichiometric ratio of A to B is 3:1, we can find the change in concentration of B: $$ \Delta[\mathrm{B}] = \frac{1}{3}(\Delta[\mathrm{A}]) = -\frac{1}{3}(5.16 \times 10^{-5} \text{ M} - 1.00 \times 10^{-4} \text{ M}) = 1.61\times10^{-5} \mathrm{M} $$ To find the concentration of B at 10.0 minutes, we can subtract this change from the initial concentration of B: $$ [\mathrm{B}]_{10} = [\mathrm{B}]_{0} - 1.61\times10^{-5} \mathrm{M} = 1.00 \mathrm{M} - 1.61\times10^{-5} \mathrm{M}= 0.999984\, \mathrm{M} $$ Therefore, the concentration of A after 10.0 minutes is \(5.16 \times 10^{-5} \mathrm{M}\), and the concentration of B is \(0.999984\, \mathrm{M}\).

Key concepts

Reaction Kinetics

Reaction kinetics explores the rate at which chemical reactions occur and the factors that affect these rates. The key feature of kinetic studies is the rate law, which expresses the rate of a reaction in terms of the concentration of each reactant. In our example, we have the reaction: \(3\mathrm{A} + \mathrm{B} + \mathrm{C} \rightarrow \mathrm{D} + \mathrm{E}\). The rate law is given by: \(-\frac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}]\). This equation tells us that the reaction rate is directly proportional to the square of the concentration of A and linearly proportional to the concentrations of B and C.
Understanding this relationship helps predict how a change in concentration affects reaction speed. For example, doubling the concentration of A would increase the rate of depletion of A fourfold (because of the squared term). Reaction kinetics is crucial for understanding chemical processes, especially in industrial and biological systems.

Half-life Calculation

Half-life in reaction kinetics is the time required for the concentration of a reactant to decrease by half. It's a useful concept because it provides insight into the dynamics of the reaction. For a complex reaction with the rate law \(-\frac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}]\), the half-life \(t_{1/2}\) is determined as \(t_{1/2} = \frac{1}{k [\mathrm{A}]_0 [\mathrm{B}]_0 [\mathrm{C}]_0}\).
This implies that the half-life depends on the initial concentrations of all reactants involved in the rate law. In our initial setup where \( k = 6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1} \) and the initial concentrations of A, B, and C are provided, we calculated \(t_{1/2}\) to be 15.4 minutes. Knowing the half-life helps in managing reaction time frames, especially in maintaining desired conditions for product yield.

Integrated Rate Law

The integrated rate law connects the concentration of reactants to time, allowing us to calculate how concentrations change throughout the duration of a reaction. For second-order reactions, such as the one involving A, the integrated rate law is: \([A]_{t}=\frac{1}{k t + 1 /[A]_{0}}\)where \( [\mathrm{A}]_{t}\) is the concentration of A at time \(t\), \(k\) is the rate constant, and \( [\mathrm{A}]_{0}\) is the initial concentration. This formula helps us to calculate the remaining concentration of A after a given time. In our example, after 10 minutes, \([A]_{t}\) was found to be \(5.16 \times 10^{-5}\, \mathrm{M}\).
The integrated rate law is valuable in scenarios where complete conversion, partial conversion, or time-dependent concentration information is necessary for the reactants.

Stoichiometry

Stoichiometry deals with the quantitative relationship between reactants and products. It relies on balanced chemical equations to calculate reactant consumption and product formation. In our reaction, two key factors are evident: for every 3 moles of A consumed, 1 mole of B is used, based on the stoichiometry \(3\mathrm{A} + \mathrm{B} + \mathrm{C} \rightarrow \mathrm{D} + \mathrm{E}\).
At a reaction duration of 10 minutes, we used stoichiometry to determine the change in concentration of B due to the consumption of A. Knowing \(\Delta[\mathrm{B}] = \frac{1}{3}(\Delta[\mathrm{A}])\), and after the calculation, \([[\mathrm{B}]_{10}\) became approximately \(0.999984\, \mathrm{M}\).
Stoichiometry ensures that all reactant and product amounts are properly accounted for, guaranteeing the mass and charge balance in chemical equations.