Integrated Rate Laws for Zero, First, and Second Order Reactions

Integrated Rate Laws for Zero, First, and Second Order Reactions

In this guide, we will cover the different integrated rate laws for zero, first, and second order reactions, providing derivations, definitions of important terms, and examples to illustrate each type of reaction.

Zero Order Reactions

  • Plotting and Slope:

    • Time is represented on the x-axis.

    • Concentration of the reactant is plotted on the y-axis.

    • The resulting plot shows a linear relationship with a negative slope.

    • Equation:

    • Standard form:

      • y = mx + b where

      • y = Concentration of reactant A ([A])

      • m = slope (negative)

      • x = time (t)

      • b = y-intercept (initial concentration [A]0)

    • The slope of the line equals the negative rate constant:

    • slope = -k

  • Half-Life Definition:

    • The half-life (T1/2) is defined as the time required for the concentration of a reactant to decrease to half its initial concentration.

  • Derivation of the Half-Life Equation:

    1. Start with the zero-order rate equation:

    • [A] = [A]0 - k * t

    1. At half-life, the concentration [A] equals 1/2[A]0:

    • 1/2[A]0 = [A]0 - k * T1/2

    1. Rearranging gives:

    • k * T1/2 = [A]0 - 1/2[A]0 = 1/2[A]0

    1. Solving for half-life yields:

    • T1/2 = [A]0 / (2k)

      • Conclusion: The half-life for a zero-order reaction is directly proportional to the initial concentration ([A]0).

First Order Reactions

  • Plotting and Slope:

    • Time is represented on the x-axis.

    • Natural log of the concentration (ln([A])) is plotted on the y-axis.

    • The plot yields a linear relationship with a negative slope.

    • Equation:

    • Standard form:

      • y = mx + b where

      • y = ln([A])

      • m = -k

      • x = t

      • b = ln([A]0)

  • Half-Life Definition:

    • For first order reactions, half-life (T1/2) is defined as the time required for the concentration to decrease to half its initial amount.

  • Derivation of the Half-Life Equation:

    1. Start with the integrated rate law:

    • ln([A]) = ln([A]0) - k * t

    1. At half-life, [A] becomes 1/2[A]0:

    • ln(1/2[A]0) = ln([A]0) - k * T1/2

    1. Utilizing properties of logarithms:

    • ln(1/2) + ln([A]0) = ln([A]0) - k * T1/2

    1. Rearranging gives:

    • k * T1/2 = -ln(1/2)

    1. Solving yields:

    • T1/2 = ln(2) / k

      • Note: The half-life for first order reactions is independent of the initial concentration.

Second Order Reactions

  • Plotting and Slope:

    • Time is represented on the x-axis.

    • The plot of the inverse of the concentration of A (1/[A]) is represented on the y-axis.

    • This results in a linear plot with a positive slope.

    • Equation:

    • Standard form:

      • y = mx + b where

      • y = 1/[A]

      • m = k

      • x = t

      • b = 1/[A]0

  • Half-Life Definition:

    • For second order reactions, the half-life (T1/2) indicates the time to reduce the concentration to half the initial value.

  • Derivation of the Half-Life Equation:

    1. Start with the rate law:

    • 1/[A] = k * t + 1/[A]0

    1. Set [A] to 1/2[A]0:

    • 1/(1/2[A]0) = k * T1/2 + 1/[A]0

    1. Simplifying the left side gives:

    • 2/[A]0 = k * T1/2 + 1/[A]0

    1. Rearranging results in:

    • k * T1/2 = 2/[A]0 - 1/[A]0 = 1/[A]0

    1. Solving for half-life yields:

    • T1/2 = 1 / (k * [A]0)

      • The half-life for second order reactions is inversely proportional to the initial concentration.

Examples of Each Reaction Type

Zero Order Example

  • Given:

    • Initial concentration [X]0 = 1.5 M

    • Rate constant k = 0.15 M/min

    • Time t = 60 min

  • Calculate remaining concentration:

    • Use the zero-order equation:

    • [A] = -kt + [A]0

    • [A] = - (0.15 M/min * 60 min) + 1.5 M

    • [A] = -9.0 + 1.5 = 0.6 M

  • Half-Life Calculation:

    • T1/2 = [A]0 / (2k) = 1.5 / (2 * 0.15) = 50 min

First Order Example

  • Given:

    • Initial concentration [A]0 = 0.25 M

    • Rate constant k = 3.05 * 10^-4 s^-1

    • Time t = 120 s

  • Calculate remaining concentration:

    • Use the first order equation:

    • ln([A]) = ln([A]0) - kt

    • ln([A]) = ln(0.25) - (3.05 * 10^-4 s^-1 * 120 s)

    • Solve:

      • ln([A]) = -1.386 - (-0.0366) = -1.423

    • Remaining concentration: [A] = e^-1.423 M
      -> 0.241 M

Second Order Example

  • Given:

    • Concentration after time [A] = 0.2 M

    • Rate constant k = 2.4 * 10^-3 M^-1 s^-1

    • Time t = 900 s

  • Calculate initial concentration [A]0:

    • Use the second order equation:

    • 1/[A] = kt + 1/[A]0

    • 1/0.2 = (2.4 * 10^-3 M^-1 s^-1 * 900 s) + 1/[A]0

    • Rearranging gives:

      • 1/[A]0 -> 5 - 2.16 -> A0 = 0.352 M

  • Half-Life for this Reaction:

    • T1/2 = 1 / (k * [A]0) = 1 / ((2.4 * 10^-3)(0.352))
      -> 1200 s