Integrated Rate Laws and Reaction Half-Lives

Integrated Rate Laws [15.4]

• Obtained by integration of the differential rate law.
• Consider the reaction: $aA \rightarrow Products$
• Rate is given by: $Rate = -\frac{d[A]}{dt} = k[A]^M$
• Rearrange the rate equation:
  $\frac{d[A]}{[A]^M} = -k dt$
• Integrate from initial concentration $[A]<em>0$ to concentration at time t $[A]</em>t$:
  $\int<em>{[A]</em>0}^{[A]<em>t} \frac{d[A]}{[A]^M} = -k \int</em>0^t dt$
• 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: $aA \rightarrow Products$
• Rate is independent of reactant concentration: $Rate = -\frac{d[A]}{dt} = k$
• Integrate the rate law:
  $\int<em>{[A]</em>0}^{[A]<em>t} d[A] = -k \int</em>0^t dt$
• This yields: $[A]<em>t - [A]</em>0 = -kt$
• Therefore, the integrated rate law is: $[A]<em>t = -kt + [A]</em>0$
• This has the form of a linear equation: $y = mx + b$, where:
  • $y = [A]_t$
  • $x = t$
  • $m = -k$ (slope)
  • $b = [A]_0$ (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: $aA \rightarrow Products$
• Rate depends on the concentration of A: $Rate = -\frac{d[A]}{dt} = k[A]$
• Rearrange and integrate:
  $\int<em>{[A]</em>0}^{[A]<em>t} \frac{d[A]}{[A]} = -k \int</em>0^t dt$
• This gives: $ln[A]<em>t - ln[A]</em>0 = -kt$
• The integrated rate law is: $ln[A]<em>t = -kt + ln[A]</em>0$
• This also has the form of a linear equation: $y = mx + b$, where:
  • $y = ln[A]_t$
  • $x = t$
  • $m = -k$ (slope)
  • $b = ln[A]_0$ (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: $aA \rightarrow Products$
• Rate depends on the square of the concentration of A: $Rate = -\frac{d[A]}{dt} = k[A]^2$
• Rearrange and integrate:
  $\int<em>{[A]</em>0}^{[A]<em>t} \frac{d[A]}{[A]^2} = -k \int</em>0^t dt$
• This yields: $\frac{1}{[A]<em>t} - \frac{1}{[A]</em>0} = kt$
• The integrated rate law is: $\frac{1}{[A]<em>t} = kt + \frac{1}{[A]</em>0}$
• Linear equation form: $y = mx + b$, where:
  • $y = \frac{1}{[A]_t}$
  • $x = t$
  • $m = k$ (slope)
  • $b = \frac{1}{[A]_0}$ (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: $2O<em>3(g) \rightarrow 3O</em>2(g)$
• Rate law: $Rate = -\frac{d[O<em>3]}{dt} = k[O</em>3]^2$
• Second-order integrated rate law:
  $\frac{1}{[O<em>3]</em>t} = \frac{1}{[O<em>3]</em>0} + kt$
• Convert time to seconds: 5.50 min * 60 sec/min = 330 sec
• Substitute values:
  $\frac{1}{[O<em>3]</em>t} = \frac{1}{0.275 M} + (0.0140 M^{-1}s^{-1})(330 s)$
  $\frac{1}{[O<em>3]</em>t} = 3.636 M^{-1} + 4.62 M^{-1} = 8.26 M^{-1}$
• Solve for $[O<em>3]</em>t$:
  $[O<em>3]</em>t = \frac{1}{8.26 M^{-1}} = 0.121 M$

Integrated Rate Law Problem 2

• Problem: Determine the rate law and rate constant for the reaction $2A \rightarrow B + C$ 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}$</li> <li>Rate law:$Rate = -\frac{d[A]}{dt} = k[A]$</li> <li>Rate constant:$k = 0.050 s^{-1}$</li> </ul> <h3 id="reactionhalflives154">Reaction Half-Lives [15.4]</h3> <ul> <li>Consider the generic reaction:$aA \rightarrow bB$</li> <li>The half-life ($t_{1/2}$) is the time it takes for half of the reactant A to be consumed.</li> <li>We will consider$t_{1/2}$for three cases: 0th order, 1st order, and 2nd order.</li> <li>Note: At$t = t{1/2}$,$[A]t = \frac{1}{2}[A]_0$</li> </ul> <h3 id="reactionhalflivesformulas">Reaction Half-Lives: Formulas</h3> <ul> <li><strong>0th Order</strong><ul> <li>Start with the integrated rate law:$[A]t = -kt + [A]0$</li> <li>Substitute$[A]t = \frac{1}{2}[A]0$and$t = t{1/2}$:$\frac{1}{2}[A]0 = -kt{1/2} + [A]0$</li> <li>Solve for$t{1/2}$:$t{1/2} = \frac{[A]_0}{2k}$</li></ul></li> <li><strong>1st Order</strong><ul> <li>Start with the integrated rate law:$ln[A]t = -kt + ln[A]0$</li> <li>Substitute$[A]t = \frac{1}{2}[A]0$and$t = t{1/2}$:$ln(\frac{1}{2}[A]0) = -kt{1/2} + ln[A]0$</li> <li>Rearrange:$ln(\frac{[A]0}{2[A]0}) = -kt_{1/2}$</li> <li>Simplify:$ln(\frac{1}{2}) = -kt_{1/2}$</li> <li>Solve for$t{1/2}$:$t{1/2} = \frac{-ln(0.5)}{k} = \frac{0.693}{k}$</li></ul></li> <li><strong>2nd Order</strong><ul> <li>Start with the integrated rate law:$\frac{1}{[A]t} = kt + \frac{1}{[A]0}$</li> <li>Substitute$[A]t = \frac{1}{2}[A]0$and$t = t{1/2}$:$\frac{1}{\frac{1}{2}[A]0} = kt{1/2} + \frac{1}{[A]0}$</li> <li>Simplify:$\frac{2}{[A]0} = kt{1/2} + \frac{1}{[A]_0}$</li> <li>Solve for$t{1/2}$:$t{1/2} = \frac{1}{k[A]_0}$</li></ul></li> </ul> <h3 id="halflifeexampleproblem">Half Life Example Problem</h3> <ul> <li>Problem: An environmental toxin is spilled into a lake. Safe concentration is below 3.5 ppm. Initial concentration is 17 ppm. The toxin decomposes with a half-life of 18 months, obeying first-order kinetics. How long until the lake is safe?</li> <li>Given:<ul> <li>$[A]_t = 3.5 ppm$</li> <li>$[A]_0 = 17 ppm$</li> <li>$t_{1/2} = 18 months$</li></ul></li> <li>First-order integrated rate law:$ln[A]t = -kt + ln[A]0$</li> <li>Calculate k from the half-life:<br />$k = \frac{-ln(0.5)}{t_{1/2}} = \frac{-ln(0.5)}{18 months} = 0.0385 months^{-1}$</li> <li>Rearrange the integrated rate law to solve for t:<br />$t = \frac{ln[A]t - ln[A]0}{-k} = \frac{ln(3.5 ppm) - ln(17 ppm)}{-0.0385 months^{-1}}$<br />$t = \frac{-ln(\frac{3.5}{17})}{0.0385 months^{-1}} = \frac{-ln(0.206)}{0.0385 months^{-1}} = 41 months$</li> </ul> <h3 id="summaryofratelawstable156">Summary of Rate Laws Table 15.6</h3> <ul> <li>For the reaction:$aA \rightarrow Products$</li> <li><strong>0th Order</strong><ul> <li>Rate Law:$Rate = -\frac{d[A]}{dt} = k$</li> <li>Integrated Rate Law:$[A]t = -kt + [A]0$</li> <li>Plot for k: [A] vs t, slope = -k</li> <li>Half-life:$t{1/2} = \frac{[A]0}{2k}$</li></ul></li> <li><strong>1st Order</strong><ul> <li>Rate Law:$Rate = k[A]$</li> <li>Integrated Rate Law:$ln[A]t = -kt + ln[A]0$</li> <li>Plot for k: ln[A] vs t, slope = -k</li> <li>Half-life:$t_{1/2} = \frac{-ln(0.5)}{k} = \frac{0.693}{k}$</li></ul></li> <li><strong>2nd Order</strong><ul> <li>Rate Law:$Rate = k[A]^2$</li> <li>Integrated Rate Law:$\frac{1}{[A]t} = kt + \frac{1}{[A]0}$</li> <li>Plot for k: 1/[A] vs t, slope = k</li> <li>Half-life:$t{1/2} = \frac{1}{k[A]0}$$