KAE_Integrated Rate Laws Lesson

Integrated Rate Laws

Often scientists need more information that just the rate of a reaction. For example, a scientist manufacturing a chemical might want to know the concentration of the reactants and products at a specific point in time so that she could decide when it would be profitable to collect that chemical. It might also be important to know when the concentration of a reactant has dropped below a certain level so that the reactant may be replenished. Or once the concentration of a reactant has dropped to a certain level, a hazardous situation could be considered safe after a certain amount of time. So, knowing the relationship between concentration and time is important. This relationship is known as the integrated rate law. You do not need to know the derivation of the integrated rate laws but might be interested to know that they are calculus based (think of integrals). The specific integrated rate law depends on the order of the reaction.

In a first-order reaction, the relationship between the concentration of A and time is shown below:

CONCENTRATION AND TIME
1st Order Reaction
In[A], kt+In[A]。 ==
[A] = concentration of A at time t
[A]o initial concentration of A at time zero

You may see this equation in different forms. For example, $LaTeX: \frac{[A]_0}{[A]_t}=kt$ $\frac{[A]_0}{[A]_t}=kt$ . Keep this equation in mind as well. It might be an easier form to use, depending upon the information given in the problem.

The equation on the blue above is written in the form of a straight line.

$LaTeX: In[A]_t=-kt+ln[A]_0 \\ \hspace{5mm}\updownarrow \hspace{13mm}\updownarrow \updownarrow \hspace{10mm}\updownarrow \\ \hspace{5mm} y\hspace{5mm} = \hspace{1mm} mx \hspace{3mm} + b$ $In[A]_t=-kt+ln[A]_0 \\ \hspace{5mm}\updownarrow \hspace{13mm}\updownarrow \updownarrow \hspace{10mm}\updownarrow \\ \hspace{5mm} y\hspace{5mm} = \hspace{1mm} mx \hspace{3mm} + b$

With the equation in this format, it is easy to see that the graph with the ln[A]_t plotted on the y-axis and t plotted on the x-axis will produce a slope of -k.

Below are graphs of the decomposition of N₂O₅ over time (at 45°C)

(a) A graph of concentration versus time for the decomposition

(b) A straight line is obtained from a logarithm versus time plot. The slope is the
negative rate constant.

image of two graphs; one with a downward sloping curve; the other with a straight line:
the slope of this line is a negative of the rate constant

An equation relating concentration and time can also be written for a 2^nd order reaction. This equation and graph information for both 1^st and 2^nd order reactions is summarized below.

1st-order
General Equation
rate = k[A]
Concentration & Time Equation
In[A]t=kt + In[A]o
Graph
Slope
In[A]t vs. t
Slope = - k

2nd-order
General Equation
rate = k[A]²
Concentration & Time Equation
1/[A]t = kt + 1/[A]o
Graph
Slope
1/[A]t vs. t
Slope = k

The decomposition of HI is a 2^nd order reaction. Here is a graph of 1/[HI] versus time. The slope of this line gives you the rate constant.

image of graph with straight, positive line; the slope of this line is the rate constant, k

Half Life

Half-life, t _½, is a convenient indicator of the rate of a reaction. The more rapid a reaction occurs, the shorter the half-life will be . Half-life is the amount of time required for half of a reactant to disappear.

You may have calculated half-life in another class simply by cutting the amount of reactant in half for each number of half-lives that have passed. You can still use that process, but only for a 1^st order reaction. Notice in the table below that in a first-order reaction the half-life does not depend on the initial concentration of the reactant. Radioactive decay is a common first-order reaction. This is why you can use the simple process you have done before. The following equations show how to calculate half-life using the rate law constant, k. The formula for calculating half-life depends on the overall order of the reaction.

t½
0-order t½ = [Ao]/2k
1st-order t½ = In 2/k
2nd-order t½= 1/k[A]o

$Example: lodine-131 is a radioactive isotope with a half-life of 8.0 days. After 32 days, what fraction of the original sample is present? We do not need to know the initial concentration to solve this problem, since it is a radioactive decay reaction (first order reaction). To determine the number of half- lives, divide the number of days by the half-life. 32/8 = 4 half-lives The fraction that remains after a number of half-lives is given by the (2)", where n is the number of half-lives that has passed. n = 4, so the fraction that remains is (%) = 1/16 Or, we can use the equations above to solve this problem. In 2 t=will allow us to solve for k. 8.0 days = In 2 k = 0.0866 1/day Then, we can use the equation relating concentration and time to solve for the ratio between [A] and [A]. [4] In = kt [4], =0.0866(32) In (4) [4], In = 2.7712 [4]。 2.7712 [4] [4], -=16 This shows that the initial concentration is 16 times larger than after 4 half-lives have passed. This is the same as the 1/16 answer we found above. This second method is longer to solve, so you may choose to use the first method when appropriate.$

Activation Energy

The Arrhenius equation relates the activation energy to the rate constant. It is written as:

$LaTeX: k=Ae^{-Ea/RT}$ $k=Ae^{-Ea/RT}$

Where k is the rate constant, A is the proportionality constant, also known as the frequency factor, e is the base of the natural logarithm system, and T is the Kelvin temperature. R is the gas constant expressed as 8.314 J mol^-1 K^-1. This equation can be rearranged algebraically to:

$LaTeX: ln(\frac{k_2}{k_1})=\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$ $ln(\frac{k_2}{k_1})=\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

Calculations involving the Arrhenius equation are beyond the scope of this course and the AP Exam. But, know that this equation can be used to summarize experiments on the temperature dependence of the rate of an elementary reaction and to interpret this dependence in terms of the activation energy needed to reach the transition state.

Remember to work on the module practice problems as you complete each section of content.

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