Question

This reaction was monitored as a function of time: $$ \mathrm{AB} \longrightarrow \mathrm{A}+\mathrm{B} $$ A plot of \(1 /[\mathrm{AB}]\) versus time yields a straight line with a slope of \(+0.55 / \mathrm{M} \cdot \mathrm{s}\) a. What is the value of the rate constant ( \(k\) ) for this reaction at this temperature? b. Write the rate law for the reaction. c. What is the half-life when the initial concentration is \(0.55 \mathrm{M} ?\) d. If the initial concentration of AB is \(0.250 \mathrm{M}\) and the reaction mixture initially contains no products, what are the concentrations of A and B after 75 s?

Short answer

a. Rate constant, k = 0.55 M^{-1}s^{-1}. b. Rate law: rate = k[AB]. c. Half-life, t_{1/2} = ln(2)/k = 1.26s. d. Concentration of A and B after 75s is approx. 0.22 M.

Step-by-step solution

Identify the order of the reaction

Since a plot of the inverse concentration of AB versus time yields a straight line, this indicates a first-order reaction, where the rate of the reaction is directly proportional to the concentration of AB.

Determine the rate constant (k)

The slope of the straight line from the given plot is actually the rate constant for a first-order reaction. Therefore, the rate constant, k, is equal to 0.55 M^{-1}s^{-1}.

Write the rate law

The rate law for a first-order reaction has the form: rate = k*[AB].

Calculate the half-life for the initial concentration

The half-life for a first-order reaction is given by the formula t_{1/2} = ln(2)/k. Substitute k = 0.55 M^{-1}s^{-1} into this formula to calculate the half-life.

Find the concentrations of A and B after 75 s

For a first-order reaction, the concentration of AB at time t can be found using the integrated rate law: ln([AB]_0/[AB]) = kt. Since the initial concentration [AB]_0 is 0.250 M and k is known, one can calculate [AB] at t=75 s. The concentration of A and B will each be equal to the change in concentration of AB, as they are produced in a 1:1 molar ratio from AB.

Key concepts

Rate Constant

In the realm of chemical kinetics, understanding the rate constant is essential. The rate constant, usually denoted by the symbol \( k \), is a measurement that indicates how fast a reaction proceeds. It's unique for each chemical reaction and is influenced by various factors such as the nature of reactants, temperature, and the presence of a catalyst.

For the given first-order reaction \( \mathrm{AB} \longrightarrow \mathrm{A}+\mathrm{B} \), the rate constant can be determined from the slope of the plot of \(1 / [\mathrm{AB}]\) versus time. The slope itself represents the rate constant, and in this case, it is \(0.55 \, \mathrm{M}^{-1}\mathrm{s}^{-1}\). This numerical value encapsulates the essence of the reaction's rate, enabling us to predict how quickly reactants are transformed into products under specified conditions.

Rate Law

The rate law is a mathematical expression that links the rate of a chemical reaction to the concentration of its reactants. It defines the dependence of the reaction rate on the concentration of the reactant(s) involved. In its general form, the rate law is given by the equation \(\text{rate} = k [\text{reactant}]^{\text{order}}\), where \(k\) is the rate constant, and the exponent signifies the order of the reaction with respect to that reactant.

For our particular first-order reaction, the rate law simplifies to \(\text{rate} = k[\mathrm{AB}]\). This tells us that the rate at which \(\mathrm{AB}\) breaks down into \(\mathrm{A}\) and \(\mathrm{B}\) is directly proportional to its concentration at any given moment. Accordingly, if the concentration of \(\mathrm{AB}\) were to double, the rate at which it reacts would also double, emphasizing the linear relationship inherent in first-order reactions.

Reaction Half-life

The half-life of a reaction, represented as \( t_{1/2} \), is the timeframe required for half of the reactant concentration to be consumed in a chemical reaction. For a first-order reaction, the half-life is remarkably constant and is not affected by the initial concentration of the reactant, distinguishing it from other reaction orders.

In the context of the given problem, the half-life can be calculated using the equation \( t_{1/2} = \frac{\ln(2)}{k} \), where \(k\) is the rate constant. Using the determined rate constant of \(0.55 \, \mathrm{M}^{-1}\mathrm{s}^{-1}\), the half-life for when the initial concentration is \(0.55 \, \mathrm{M}\) can be computed. This predictable and steady half-life is a hallmark of first-order kinetics and provides useful insight when planning or analyzing experiments.

Integrated Rate Law

The integrated rate law is a powerful tool that allows chemists to link the concentrations of reactants to time for a given reaction order. It is the result of integrating the rate law over time and provides an equation that can be used to calculate the concentration of a reactant at any time during a reaction. For first-order reactions, the integrated rate law takes the form \(\ln([\mathrm{AB}]_{0}/[\mathrm{AB}]) = kt\), where \(\ln\) represents the natural logarithm, \(k\) is the rate constant, \(t\) is time, and \( [\mathrm{AB}]_{0} \) and \( [\mathrm{AB}]\) are the initial and current concentrations of the reactant, respectively.

Implementing this law with the given data, if the initial concentration of \(\mathrm{AB}\) is \(0.250 \, \mathrm{M}\) and we calculate at \(t = 75 \, \mathrm{s}\), we can find the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\). Considering \(\mathrm{A}\) and \(\mathrm{B}\) are formed in a 1:1 ratio from the decomposition of \(\mathrm{AB}\), we can infer that each product's concentration will be equal to the decrease in \(\mathrm{AB}\)'s concentration over the course of the reaction. This again reflects the balanced nature of the reaction equation and supports the stoichiometric understanding of the reaction's progress.