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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Kinetics > Reaction Rates > First-Order Reactions

First-Order Reactions

A first-order reaction is a reaction that proceeds at a rate that depends linearly only on one reactant concentration. The rate at which a reactant is consumed in a first-order process is proportional to its concentration at that time.

The Differential Rate Law

A first-order reaction is a reaction that proceeds at a rate that depends linearly only on one reactant concentration. The differential equation describing first order kinetics is:

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

\( Rate\) is the reaction rate (in unites of molar/time) and \( k\) is the reaction rate coefficient (in units of 1/time). However, the units can vary with other order reactions. These differential equations are separable, which simplifies the solutions.

The Integral Rate Law

First, write the differential form of the rate law.

\[ Rate = - \dfrac{d[A]}{dt} = k[A] \tag{2}\]

Rearrange

\[ \dfrac{d[A]}{[A]} = - k\,dt \tag{3}\]

Second, integrate both sides of the equation.

\[ \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]} = -\int_{t_o}^{t} k\, dt \tag{4a}\]

\[ \int_{[A]_{o}}^{[A]} \dfrac{1}{[A]} d[A] = -\int_{t_o}^{t} k\, dt \tag{4b}\]

Recall from calculus

\[ \int  \dfrac{1}{x} = \ln(x) \tag{5}\]

Upon integration, we get

\[ \ln[A] - \ln[A]_o = -kt \tag{6}\]

Rearrange to solve for \( [A] \) and we get one form of the rate law

\[ \ln[A] = \ln[A]_o - kt \tag{7}\]

We can rearrange the equation above to:

\[ \ln [A] = -kt + \ln [A]_o \tag{8}\]

Recall from Algebra, that \(y=mx +b\) is the equation of a straight line so :

\[ \ln [A] = -kt + \ln [A]_o \tag{9}\]

is a straight line with slope (\(m\):

\[mx=-kt \tag{10}\]

and y-intercept (\(b\)):

\[b=\ln [A]_o \tag{11}\]

Now, recall from the laws of logarithms that 

\[ \ln {\left(\dfrac{[A]_t}{ [A]_o}\right)}= -kt \tag{12}\]

is at the time \(t\) with its final concentration of \(A\) and \([A]_o\) is at time \(0\) and it is at its initial concentration of \([A]_o\) and \(k\) is  the rate constant.

   

The "e" in the exponential term is of course the base of the natural logarithms, and the negative sign in its exponent means that the value of this term diminishes as \(t\) increases, as we would expect for any kind of a decay process.

A more convenient form of the integrated rate law is obtained by taking the natural logarithm of both sides:

\[\ln [A] = –kt + \ln [A]_o\tag{13}\]

This has the form of an equation for a straight line

\[y = mx + b \tag{14}\]

in which the slope \(m\) corresponds to the rate constant \(k\). This means that, for a first-order reaction, a plot of \(\ln [A]\) as a function of time gives a straight line with a slope of \(–k\). Since, the logarithms of numbers do not have any units, the product of -kt does not have units as well. This concludes that unit of k in a first order of reaction must be time-1. Examples of time-1 would be s-1 or min-1. Thus, the equation of a straight line is applicable to represent:

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

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, \(e\)

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

Taking the natural log of both sides of the equation, we get the second form of the rate law

\[ [A] = [A]_{o}e^{- kt}\tag{17}\]

The integrated forms of the rate law allow us 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 \(-k\). For more information on differential and integrated 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 = ex so efficiently describes such changes stems from the remarkable property that dy/dx = ex; that is, ex is its own derivative, so the rate of change of y is 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.

Image04.jpg         Image05.jpg

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

First order.jpg

Half-live of First order Reactions

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

\[ [A] = \dfrac{1}{2} [A]_o \]

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

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

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

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

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

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

which tells us that the half-life of a first-order reaction is a constant. This means that 100,000 molecules of a reactant will be reduced to 50,000 in the same time interval needed for ten molecules to be reduced to five.

 

 

Example 5

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

SOLUTION

From the above equation, k = –0.693/(600 s) = 0.00115 s–1

The decay of radioactive nuclei is always a first-order process.

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 for \([A]\) to decrease from 1 M to 0.5 M, it takes much time as it does for \([A]\) to decrease from 0.1 M to 0.05 M. It should also be clear that the rate constant and the half life of a first-order process are inversely related.

Example: Determining Half life

If 3.0 g of substance \(A\) decomposes for \(36\, min\), the mass of \(A\) remaining unreacted 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 problems:

  • Approach #1: Recognize that the final concentration of \(A\) is \(\frac{1}{8}\) of the final concentration and hence three half lives (\(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\)) have elapsed during this reaction. \[t_{1/2} = \dfrac{36\, \text{min}}{3}= 12 \; \text{min}\]
  • Approach #2: "The brute force approach" involves solving for \(k\) from the integral rate law equation (eq. 12 or 17) and then relating \(k\) to the \(t_{1/2}\) via equation 20.

\[\dfrac{[A]_t}{[A]_o} =e^{-k\,t}\]

\[k= -\dfrac{\ln \dfrac{[A]_t}{[A]_o}}{t} = -\dfrac{\ln \dfrac{0.375\, g}{3\, g}}{36\, \text{min}} = 0.0578 \, \text{min}^{-1}\]

\[t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{0.1268 \, \text{min}^{-1}} \approx 12\, \text{min}\]

The first approach is considerably faster (if you can recognize how many half lives the reactions have evolved).

Practice Problems

Calculate the half-life of the reactions below:

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

Determine the percent 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. Use the value of k above. 
  2. The time for the concentration to decompose is 450 s after the reaction begins. Use the value of k above.

Solutions

Use the half life reaction that contains initial concentration and final concentration. Plug in the appropriate variables and you should get:

  1. 17.2 min
  2. 9.67 min
  3. 5.75 min
  4. Rearranging Eq. 17 to solve for the  \([H_2O_2]_t/[H_2O_2]_0\) ratio​ \[\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt}\] This is a simple plug and play application once you have identified this equation. \[\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (600 \, s)}\] \[\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = 0.9629\] So 100-96.3=3.71% of the hydrogen peroxide has decayd by 600 s.
  5. Rearranging Eq. 17 to solve for the  \([H_2O_2]_t/[H_2O_2]_0\) ratio​ \[\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt}\] This is a simple plug and play application once you have identified this equation. \[\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (450 \, s)}\] \[\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = 0.9720\] So 100-96.3=2.8% of the hydrogen peroxide has decayd by 450 s.

Summary

The kinetics of any reaction depend on the reaction mechanism, or rate law, and the initial conditions. If we assume for the reaction A -> products, there is an initial concentration of reactant of [A]0 at time t=0, and the rate law is an integral order in A, then we can summarize the kinetics of the the first order reaction as follows:

1st.jpg

References

  1. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications 9th Ed. New Jersey: Pearson Education Inc. 2007.
  2. http://en.wikipedia.org/wiki/Half-life

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Last Modified
19:50, 26 May 2014

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