What is the slope of the graph for a first order reaction?

What is the slope of the graph for a first order reaction?

• A. -k
• B. +k
• C. 0
• D. k/2

Answer: (A)

For a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction can be expressed as: \[ \ln[A] = -kt + \ln[A_0] \] In this equation, \([A]\) is the concentration of the reactant at time \(t\), \([A_0]\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time. This equation is in the form of the equation of a straight line \(y = mx + b\), where \(y\) is \(\ln[A]\), \(m\) is the slope, \(x\) is \(t\), and \(b\) is \(\ln[A_0]\). From the rearranged equation, it’s clear that the slope of the line is given by \(-k\). Thus, for a plot of \(\ln[A]\) versus time, the slope will be negative because the concentration of the reactant decreases over time, which reflects the consumption of the reactant as the reaction proceeds. Therefore, the correct answer emphasizes that for a first-order reaction the slope of the

For a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction can be expressed as:

[ \ln[A] = -kt + \ln[A_0] ]

In this equation, ([A]) is the concentration of the reactant at time (t), ([A_0]) is the initial concentration, (k) is the rate constant, and (t) is time. This equation is in the form of the equation of a straight line (y = mx + b), where (y) is (\ln[A]), (m) is the slope, (x) is (t), and (b) is (\ln[A_0]).

From the rearranged equation, it’s clear that the slope of the line is given by (-k). Thus, for a plot of (\ln[A]) versus time, the slope will be negative because the concentration of the reactant decreases over time, which reflects the consumption of the reactant as the reaction proceeds. Therefore, the correct answer emphasizes that for a first-order reaction the slope of the