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. b. Write the rate law for the reaction.

검증된 단계별 안내

Identify the order of the reaction by analyzing the plot. A straight line in a plot of 1/[AB] versus time indicates a second-order reaction.

For a second-order reaction, the rate law is expressed as: rate = k[AB]^2.

The slope of the line in 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 represents the rate constant k.

Therefore, the rate law for the reaction is: rate = 0.55 [AB]^2.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Rate Law

The rate law of a chemical reaction expresses the relationship between the rate of the reaction and the concentration of its reactants. It is typically formulated as rate = k[A]^m[B]^n, where k is the rate constant, and m and n are the orders of the reaction with respect to reactants A and B. Understanding the rate law is essential for predicting how changes in concentration affect the reaction rate.

Order of Reaction

The order of a reaction refers to the power to which the concentration of a reactant is raised in the rate law. It indicates how the rate of reaction is affected by the concentration of that reactant. In this case, since the plot of 1/[AB] versus time is linear, it suggests that the reaction is second-order with respect to AB, meaning the rate is proportional to the square of the concentration of AB.

Integrated Rate Laws

Integrated rate laws relate the concentration of reactants to time and are derived from the differential rate laws. For a second-order reaction, the integrated rate law is given by 1/[AB] = kt + 1/[AB]0, where [AB]0 is the initial concentration. The linear relationship observed in the plot indicates that the reaction follows this integrated form, allowing us to determine the rate constant from the slope.