Integrated Rate Laws and Reaction Half-Lives

Integrated Rate Laws [15.4]

  • Obtained by integration of the differential rate law.
  • Consider the reaction: aA→ProductsaA \rightarrow Products
  • Rate is given by: Rate=−d[A]dt=k[A]MRate = -\frac{d[A]}{dt} = k[A]^M
  • Rearrange the rate equation:
    d[A][A]M=−kdt\frac{d[A]}{[A]^M} = -k dt
  • Integrate from initial concentration [A]<em>0[A]<em>0 to concentration at time t [A]</em>t[A]</em>t:
    ∫<em>[A]</em>0[A]<em>td[A][A]M=−k∫</em>0tdt\int<em>{[A]</em>0}^{[A]<em>t} \frac{d[A]}{[A]^M} = -k \int</em>0^t dt
  • Solutions provide insight into:
    • Concentration of A at a specific time.
    • Time required to reach a certain concentration of A.
    • Initial rates.
  • Note: Initial rates can be difficult to obtain accurately; concentration data over time is often easier to measure.

Integrated 0th Order Rate Law

  • For a zero-order reaction: aA→ProductsaA \rightarrow Products
  • Rate is independent of reactant concentration: Rate=−d[A]dt=kRate = -\frac{d[A]}{dt} = k
  • Integrate the rate law:
    ∫<em>[A]</em>0[A]<em>td[A]=−k∫</em>0tdt\int<em>{[A]</em>0}^{[A]<em>t} d[A] = -k \int</em>0^t dt
  • This yields: [A]<em>t−[A]</em>0=−kt[A]<em>t - [A]</em>0 = -kt
  • Therefore, the integrated rate law is: [A]<em>t=−kt+[A]</em>0[A]<em>t = -kt + [A]</em>0
  • This has the form of a linear equation: y=mx+by = mx + b, where:
    • y=[A]ty = [A]_t
    • x=tx = t
    • m=−km = -k (slope)
    • b=[A]0b = [A]_0 (y-intercept)
  • Units of k for a zero-order reaction: M s$^{-1}$ or M time$^{-1}$

Integrated 1st Order Rate Law

  • For a first-order reaction: aA→ProductsaA \rightarrow Products
  • Rate depends on the concentration of A: Rate=−d[A]dt=k[A]Rate = -\frac{d[A]}{dt} = k[A]
  • Rearrange and integrate:
    ∫<em>[A]</em>0[A]<em>td[A][A]=−k∫</em>0tdt\int<em>{[A]</em>0}^{[A]<em>t} \frac{d[A]}{[A]} = -k \int</em>0^t dt
  • This gives: ln[A]<em>t−ln[A]</em>0=−ktln[A]<em>t - ln[A]</em>0 = -kt
  • The integrated rate law is: ln[A]<em>t=−kt+ln[A]</em>0ln[A]<em>t = -kt + ln[A]</em>0
  • This also has the form of a linear equation: y=mx+by = mx + b, where:
    • y=ln[A]ty = ln[A]_t
    • x=tx = t
    • m=−km = -k (slope)
    • b=ln[A]0b = ln[A]_0 (y-intercept)
  • Represented graphically as exponential decay.
  • Units of k for a first-order reaction: s$^{-1}$ or time$^{-1}$

Integrated 2nd Order Rate Law

  • For a second-order reaction: aA→ProductsaA \rightarrow Products
  • Rate depends on the square of the concentration of A: Rate=−d[A]dt=k[A]2Rate = -\frac{d[A]}{dt} = k[A]^2
  • Rearrange and integrate:
    ∫<em>[A]</em>0[A]<em>td[A][A]2=−k∫</em>0tdt\int<em>{[A]</em>0}^{[A]<em>t} \frac{d[A]}{[A]^2} = -k \int</em>0^t dt
  • This yields: 1[A]<em>t−1[A]</em>0=kt\frac{1}{[A]<em>t} - \frac{1}{[A]</em>0} = kt
  • The integrated rate law is: 1[A]<em>t=kt+1[A]</em>0\frac{1}{[A]<em>t} = kt + \frac{1}{[A]</em>0}
  • Linear equation form: y=mx+by = mx + b, where:
    • y=1[A]ty = \frac{1}{[A]_t}
    • x=tx = t
    • m=km = k (slope)
    • b=1[A]0b = \frac{1}{[A]_0} (y-intercept)
  • Units of k for a second-order reaction: M$^{-1}$s$^{-1}$ or M$^{-1}$time$^{-1}$

Integrated Rate Law Problem 1

  • Problem: Decay of ozone into dioxygen obeys second-order kinetics with k = 0.0140 M$^{-1}$s$^{-1}$. Initial ozone concentration is 0.275 M. Find the concentration after 5.50 minutes.
  • Reaction: 2O<em>3(g)→3O</em>2(g)2O<em>3(g) \rightarrow 3O</em>2(g)
  • Rate law: Rate=−d[O<em>3]dt=k[O</em>3]2Rate = -\frac{d[O<em>3]}{dt} = k[O</em>3]^2
  • Second-order integrated rate law:
    1[O<em>3]</em>t=1[O<em>3]</em>0+kt\frac{1}{[O<em>3]</em>t} = \frac{1}{[O<em>3]</em>0} + kt
  • Convert time to seconds: 5.50 min * 60 sec/min = 330 sec
  • Substitute values:
    1[O<em>3]</em>t=10.275M+(0.0140M−1s−1)(330s)\frac{1}{[O<em>3]</em>t} = \frac{1}{0.275 M} + (0.0140 M^{-1}s^{-1})(330 s)
    1[O<em>3]</em>t=3.636M−1+4.62M−1=8.26M−1\frac{1}{[O<em>3]</em>t} = 3.636 M^{-1} + 4.62 M^{-1} = 8.26 M^{-1}
  • Solve for [O<em>3]</em>t[O<em>3]</em>t:
    [O<em>3]</em>t=18.26M−1=0.121M[O<em>3]</em>t = \frac{1}{8.26 M^{-1}} = 0.121 M

Integrated Rate Law Problem 2

  • Problem: Determine the rate law and rate constant for the reaction 2A→B+C2A \rightarrow B + C given concentration vs. time data.
  • Strategy: Plot the data in different ways to determine the order of the reaction.
    • Plot [A] vs t (0th order).
    • Plot ln[A] vs t (1st order).
    • Plot 1/[A] vs t (2nd order).
  • The plot that yields a linear relationship indicates the order of the reaction.
  • Extract the slope from the linear plot to determine the rate constant k.
  • Formulate the differential rate law.
  • Data analysis:
    • Given data includes time, [A], ln([A]), and [A]^{-1}$ .
    • The plot of ln[A] vs t appears linear.
  • Since the plot of ln[A] vs time is linear, the reaction is first order.
  • For a first-order reaction, the slope of ln[A] vs t gives -k.
  • Calculate the slope:
    k = -slope = -\frac{(1.11 - 1.61)}{(10 - 0)} = -\frac{-0.50}{10} s^{-1} = 0.050 s^{-1}
  • Rate law: Rate = -\frac{d[A]}{dt} = k[A]
  • Rate constant: k = 0.050 s^{-1}

Reaction Half-Lives [15.4]

  • Consider the generic reaction: aA \rightarrow bB
  • The half-life (t_{1/2}) is the time it takes for half of the reactant A to be consumed.
  • We will consider t_{1/2} for three cases: 0th order, 1st order, and 2nd order.
  • Note: At t = t{1/2},,[A]t = \frac{1}{2}[A]_0

Reaction Half-Lives: Formulas

  • 0th Order
    • Start with the integrated rate law: [A]t = -kt + [A]0
    • Substitute [A]t = \frac{1}{2}[A]0andandt = t{1/2}::\frac{1}{2}[A]0 = -kt{1/2} + [A]0
    • Solve for t{1/2}::t{1/2} = \frac{[A]_0}{2k}
  • 1st Order
    • Start with the integrated rate law: ln[A]t = -kt + ln[A]0
    • Substitute [A]t = \frac{1}{2}[A]0andandt = t{1/2}::ln(\frac{1}{2}[A]0) = -kt{1/2} + ln[A]0
    • Rearrange: ln(\frac{[A]0}{2[A]0}) = -kt_{1/2}
    • Simplify: ln(\frac{1}{2}) = -kt_{1/2}
    • Solve for t{1/2}::t{1/2} = \frac{-ln(0.5)}{k} = \frac{0.693}{k}
  • 2nd Order
    • Start with the integrated rate law: \frac{1}{[A]t} = kt + \frac{1}{[A]0}
    • Substitute [A]t = \frac{1}{2}[A]0andandt = t{1/2}::\frac{1}{\frac{1}{2}[A]0} = kt{1/2} + \frac{1}{[A]0}
    • Simplify: \frac{2}{[A]0} = kt{1/2} + \frac{1}{[A]_0}
    • Solve for t{1/2}::t{1/2} = \frac{1}{k[A]_0}

Half Life Example Problem

  • Problem: An environmental toxin is spilled into a lake. Safe concentration is below 3.5 ppm. Initial concentration is 17 ppm. The toxin decomposes with a half-life of 18 months, obeying first-order kinetics. How long until the lake is safe?
  • Given:
    • [A]_t = 3.5 ppm
    • [A]_0 = 17 ppm
    • t_{1/2} = 18 months
  • First-order integrated rate law: ln[A]t = -kt + ln[A]0
  • Calculate k from the half-life:
    k = \frac{-ln(0.5)}{t_{1/2}} = \frac{-ln(0.5)}{18 months} = 0.0385 months^{-1}
  • Rearrange the integrated rate law to solve for t:
    t = \frac{ln[A]t - ln[A]0}{-k} = \frac{ln(3.5 ppm) - ln(17 ppm)}{-0.0385 months^{-1}}
    t = \frac{-ln(\frac{3.5}{17})}{0.0385 months^{-1}} = \frac{-ln(0.206)}{0.0385 months^{-1}} = 41 months

Summary of Rate Laws Table 15.6

  • For the reaction: aA \rightarrow Products
  • 0th Order
    • Rate Law: Rate = -\frac{d[A]}{dt} = k
    • Integrated Rate Law: [A]t = -kt + [A]0
    • Plot for k: [A] vs t, slope = -k
    • Half-life: t{1/2} = \frac{[A]0}{2k}
  • 1st Order
    • Rate Law: Rate = k[A]
    • Integrated Rate Law: ln[A]t = -kt + ln[A]0
    • Plot for k: ln[A] vs t, slope = -k
    • Half-life: t_{1/2} = \frac{-ln(0.5)}{k} = \frac{0.693}{k}
  • 2nd Order
    • Rate Law: Rate = k[A]^2
    • Integrated Rate Law: \frac{1}{[A]t} = kt + \frac{1}{[A]0}
    • Plot for k: 1/[A] vs t, slope = k
    • Half-life: t{1/2} = \frac{1}{k[A]0}$$