2.3: First-Order Reactions - Chemistry LibreTexts

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1. The Differential Representation
2. The Integral Representation
3. Graphing First-order Reactions
4. Half-lives of first order reactions
   1. Example 1: Estimated Rate Constants
   2. Example \(\PageIndex{1}\): Determining Half life
      1. Solution
   3. Exercise \(\PageIndex{2a}\)
   4. Exercise \(\PageIndex{2b}\)
5. References
6. Contributors and Attributions

A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.

The Differential Representation

Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. The differential equation describing first-order kinetics is given below:

\[ Rate = - \dfrac{d[A]}{dt} = k[A]^1 = k[A] \label{1} \]

The "rate" is the reaction rate (in units of molar/time) and \(k\) is the reaction rate coefficient (in units of 1/time). However, the units of \(k\) vary for non-first-order reactions. These differential equations are separable, which simplifies the solutions as demonstrated below.

The Integral Representation

First, write the differential form of the rate law.

\[ Rate = - \dfrac{d[A]}{dt} = k[A] \nonumber \]

Rearrange to give:

\[ \dfrac{d[A]}{[A]} = - k\,dt \nonumber \]

Second, integrate both sides of the equation.

\[ \begin{align*} \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]} &= -\int_{t_o}^{t} k\, dt \label{4a} \\[4pt] \int_{[A]_{o}}^{[A]} \dfrac{1}{[A]} d[A] &= -\int_{t_o}^{t} k\, dt \label{4b} \end{align*} \]

Recall from calculus that:

\[ \int \dfrac{1}{x} = \ln(x) \nonumber \]

Upon integration,

\[ \ln[A] - \ln[A]_o = -kt \nonumber \]

Rearrange to solve for [A] to obtain one form of the rate law:

\[ \ln[A] = \ln[A]_o - kt \nonumber \]

This can be rearranged to:

\[ \ln [A] = -kt + \ln [A]_o \nonumber \]

This can further be arranged into y=mx +b form:

\[ \ln [A] = -kt + \ln [A]_o \nonumber \]

The equation is a straight line with slope m:

\[mx=-kt \nonumber \]

and y-intercept b:

\[b=\ln [A]_o \nonumber \]

Now, recall from the laws of logarithms that

\[ \ln {\left(\dfrac{[A]_t}{ [A]_o}\right)}= -kt \nonumber \]

where [A] is the concentration at time \(t\) and \([A]_o\) is the concentration at time 0, and \(k\) is the first-order rate constant.

Because the logarithms of numbers do not have any units, the product \(-kt\) also lacks units. This concludes that unit of \(k\) in a first order of reaction must be time-1. Examples of time-1 include s-1 or min-1. Thus, the equation of a straight line is applicable:

\[ \ln [A] = -kt + \ln [A]_o.\label{15} \]

To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction.

To create another form of the rate law, raise each side of the previous equation to the exponent, \(e\):

\[ \large e^{\ln[A]} = e^{\ln[A]_o - kt} \label{16} \]

Simplifying gives the second form of the rate law:

\[ [A] = [A]_{o}e^{- kt}\label{17} \]

The integrated forms of the rate law can be used to find the population of reactant at any time after the start of the reaction. Plotting \(\ln[A]\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\). More information can be found in the article on rate laws.

This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law. Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function \(y=e^x\) so efficiently describes such changes is that dy/dx = ex; that is, ex is its own derivative, making the rate of change of \(y\) identical to its value at any point.

Graphing First-order Reactions

The following graphs represents concentration of reactants versus time for a first-order reaction.

Plotting \(\ln[A]\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\).

Half-lives of first order reactions

The half-life (\(t_{1/2}\)) is a timescale on which the initial population is decreased by half of its original value, represented by the following equation.

\[ [A] = \dfrac{1}{2} [A]_o \nonumber \]

After a period of one half-life, \(t = t_{1/2}\) and we can write

\[\dfrac{[A]_{1/2}}{[A]_o} = \dfrac{1}{2}=e^{-k\,t_{1/2}} \label{18} \]

Taking logarithms of both sides (remember that \(\ln e^x = x\)) yields

\[ \ln 0.5 = -kt\label{19} \]

Solving for the half-life, we obtain the simple relation

\[ t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{k}\label{20} \]

This indicates that the half-life of a first-order reaction is a constant.

Example 1: Estimated Rate Constants

The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant?

Solution

Use Equation 20 that relates half life to rate constant for first order reactions:

\[k = \dfrac{0.693}{600 \;s} = 0.00115 \;s^{-1} \nonumber \]

As a check, dimensional analysis can be used to confirm that this calculation generates the correct units of inverse time.

Notice that, for first-order reactions, the half-life is independent of the initial concentration of reactant, which is a unique aspect to first-order reactions. The practical implication of this is that it takes as much time for [A] to decrease from 1 M to 0.5 M as it takes for [A] to decrease from 0.1 M to 0.05 M. In addition, the rate constant and the half life of a first-order process are inversely related.

Example \(\PageIndex{1}\): Determining Half life

If 3.0 g of substance \(A\) decomposes for 36 minutes the mass of unreacted A remaining is found to be 0.375 g. What is the half life of this reaction if it follows first-order kinetics?

Solution

There are two ways to approach this problem: The "simple inspection approach" and the "brute force approach"

Approach #1: "The simple Inspection Approach"

This approach is used when one can recognize that the final concentration of \(A\) is \(\frac{1}{8}\) of the initial concentration and hence three half lives \(\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)\) have elapsed during this reaction.

\[t_{1/2} = \dfrac{36\, \text{min}}{3}= 12 \; \text{min} \nonumber \]

This approach works only when the final concentration is \(\left(\frac{1}{2}\right)^n\) that of the initial concentration, then \(n\) is the number of half lives that have elapsed. If this is not the case, then approach #2 can be used.

Approach #2: "The brute force approach"

This approach involves solving for \(k\) from the integral rate law (Equation \ref{17}) and then relating \(k\) to the \(t_{1/2}\) via Equation \ref{20}. \[\begin{align*} \dfrac{[A]_t}{[A]_o} &= e^{-k\,t} \\[4pt] k &= -\dfrac{\ln \dfrac{[A]_t}{[A]_o}}{t} \\[4pt] &= -\dfrac{\ln \dfrac{0.375\, g}{3\, g}}{36\, \text{min}} \\[4pt] &= 0.0578 \, \text{min}^{-1} \end{align*} \] Therefore, via Equation \ref{20} \[t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{0.0578 \, \text{min}^{-1}} \approx 12\, \text{min} \nonumber \]

The first approach is considerably faster (if the number of half lives evolved is apparent).

Exercise \(\PageIndex{2a}\)

Calculate the half-life of the reactions below:

1. If 4.00 g A are allowed to decompose for 40 min, the mass of A remaining undecomposed is found to be 0.80 g.
2. If 8.00 g A are allowed to decompose for 34 min, the mass of A remaining undecomposed is found to be 0.70 g.
3. If 9.00 g A are allowed to decompose for 24 min, the mass of A remaining undecomposed is found to be 0.50 g.

Answer

Use the half life reaction that contains initial concentration and final concentration. Plug in the appropriate variables and solve to obtain:

1. 17.2 min
2. 9.67 min
3. 5.75 min

Exercise \(\PageIndex{2b}\)

Determine the percent \(\ce{H2O2}\) that decomposes in the time using \(k=6.40 \times 10^{-5} s^{-1}\)

1. The time for the concentration to decompose is 600.0 s after the reaction begins.
2. The time for the concentration to decompose is 450 s after the reaction begins.

Answer

1. Rearranging Eq. 17 to solve for the \([H_2O_2]_t/[H_2O_2]_0\)ratio​ \[\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt} \nonumber \] This is a simple plug and play application once you have identified this equation. \[\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (600 \, s)} \nonumber \] \[\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = 0.9629 \nonumber \] So 100-96.3=3.71% of the hydrogen peroxide has decayed by 600 s.
2. Rearranging Eq. 17 to solve for the \([H_2O_2]_t/[H_2O_2]_0\)ratio​ \[\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt} \nonumber \] This is a simple plug and play application once you have identified this equation. \[\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (450 \, s)} \nonumber \] \[\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = 0.9720 \nonumber \] So 100-97.2=2.8% of the hydrogen peroxide has decayed by 450 s.

References

1. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications 9th Ed. New Jersey: Pearson Education Inc. 2007.

Contributors and Attributions

• Rachael Curtis (UCD), Cathy Nguyen (UCD)

• Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook

2.3: First-Order Reactions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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2. • 2.2: Factors That Affect Reaction Rates
   • 2.4: Half-lives

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