This reaction was monitored as a function of time: AB → A + B A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/Ms.
a. What is the value of the rate constant (k) for this reaction at this temperature?

Verified step by step guidance

1. Recognize that the plot of \( \frac{1}{[\text{AB}]} \) versus time indicates a second-order reaction.
2. For a second-order reaction, the integrated rate law is \( \frac{1}{[\text{AB}]} = kt + \frac{1}{[\text{AB}]_0} \), where \( k \) is the rate constant.
3. In the equation of a straight line \( y = mx + c \), the slope \( m \) corresponds to the rate constant \( k \) for a second-order reaction.
4. Given that the slope of the line is \(+0.55 \text{ M}^{-1}\text{s}^{-1}\), this value is the rate constant \( k \).
5. Therefore, the rate constant \( k \) for this reaction is \( 0.55 \text{ M}^{-1}\text{s}^{-1} \).

Key Concepts

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For a reaction of the form AB → A + B, the rate can be determined by the rate constant (k) and the concentration of AB raised to a power that reflects its order. Understanding the rate law is essential for calculating the rate constant from experimental data.

Integrated Rate Laws

Integrated rate laws relate the concentration of reactants to time. For a second-order reaction, the integrated rate law is expressed as 1/[AB] = kt + 1/[AB]₀, where k is the rate constant, t is time, and [AB]₀ is the initial concentration. The linear relationship observed in the plot of 1/[AB] versus time indicates that the reaction follows second-order kinetics.

Slope of a Line

In the context of a linear plot, the slope represents the rate constant (k) for the reaction. For the plot of 1/[AB] versus time, the slope is equal to k, which is given as +0.55/Ms in this case. This value directly provides the rate constant needed to describe the kinetics of the reaction.