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]0 to concentration at time t [A]t:
\int{[A]0}^{[A]t} \frac{d[A]}{[A]^M} = -k \int0^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.

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{[A]0}^{[A]t} d[A] = -k \int0^t dt
This yields: [A]t - [A]0 = -kt
Therefore, the integrated rate law is: [A]t = -kt + [A]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$^{-1}$

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{[A]0}^{[A]t} \frac{d[A]}{[A]} = -k \int0^t dt
This gives: ln[A]t - ln[A]0 = -kt
The integrated rate law is: ln[A]t = -kt + ln[A]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$^{-1}$

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{[A]0}^{[A]t} \frac{d[A]}{[A]^2} = -k \int0^t dt
This yields: \frac{1}{[A]t} - \frac{1}{[A]0} = kt
The integrated rate law is: \frac{1}{[A]t} = kt + \frac{1}{[A]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$^{-1}$

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: 2O3(g) \rightarrow 3O2(g)
Rate law: Rate = -\frac{d[O3]}{dt} = k[O3]^2
Second-order integrated rate law:
\frac{1}{[O3]t} = \frac{1}{[O3]0} + kt
Convert time to seconds: 5.50 min * 60 sec/min = 330 sec
Substitute values:
\frac{1}{[O3]t} = \frac{1}{0.275 M} + (0.0140 M^{-1}s^{-1})(330 s)
\frac{1}{[O3]t} = 3.636 M^{-1} + 4.62 M^{-1} = 8.26 M^{-1}
Solve for [O3]t:
[O3]t = \frac{1}{8.26 M^{-1}} = 0.121 M

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}
Rate law: Rate = -\frac{d[A]}{dt} = k[A]
Rate constant: k = 0.050 s^{-1}

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

0th Order
- Start with the integrated rate law: [A]t = -kt + [A]0
- Substitute [A]t = \frac{1}{2}[A]0 and t = 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]0 and t = 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]0 and t = 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}

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

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}$$