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. c. What is the half-life when the initial concentration is 0.55 M?

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Identify the order of the reaction. Since a plot of 1/[AB] versus time yields a straight line, the reaction is second order.

Use the second-order half-life formula: \( t_{1/2} = \frac{1}{k[A]_0} \), where \( k \) is the rate constant and \( [A]_0 \) is the initial concentration.

Recognize that the slope of the line in a second-order reaction plot is equal to the rate constant \( k \). Therefore, \( k = 0.55 \text{ M}^{-1}\text{s}^{-1} \).

Substitute the given initial concentration \( [A]_0 = 0.55 \text{ M} \) and the rate constant \( k = 0.55 \text{ M}^{-1}\text{s}^{-1} \) into the half-life formula.

Calculate the half-life using the formula: \( t_{1/2} = \frac{1}{0.55 \times 0.55} \).

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주요 개념

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Rate Laws and Reaction Order

Rate laws describe the relationship between the concentration of reactants and the rate of a chemical reaction. The order of a reaction indicates how the rate is affected by the concentration of each reactant. In this case, the reaction AB → A + B is second-order, as indicated by the linear plot of 1/[AB] versus time, which is characteristic of second-order kinetics.

Half-Life of a Reaction

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value. For second-order reactions, the half-life is dependent on the initial concentration of the reactant. The formula for the half-life (t1/2) of a second-order reaction is t1/2 = 1/(k[AB]0), where k is the rate constant and [AB]0 is the initial concentration.

Slope of the Rate Plot

In the context of reaction kinetics, the slope of the plot of 1/[AB] versus time is equal to the rate constant (k) for a second-order reaction. Given that the slope is +0.55/Ms, this value directly represents the rate constant used in calculating the half-life. Understanding how to interpret the slope is crucial for determining the kinetics of the reaction.