Reactions happen at all kinds of speeds. The combustion of natural gas can happen almost instantly, but the rusting of iron may take hours or even days.
So, why is that the case? There are two reasons: the first is the rate constant (k). Which is a unique constant that changes based on the type of reaction and the temperature. The second is the concentration of the reactant(s). The magnitude at which the concentration affects the rate is called the order. In this article, we will be diving into second-order reactions.
- This article is about second-order reactions
- First, we will look at some examples of second-order reactions
- Next we will identify the units for the rate constant
- Then we will derive the integrated rate equation for the two types of second-order reactions
- We will then graph these equations and see how we can use the graphs to calculate the rate constant
- Lastly, we will derive and use the half-life equation for second-order reactions.
Second-order Reaction Examples and Definition
Let's first define what a second-order reaction is:
A second-order reaction is a reaction whose rate is dependent on either of two cases:
- the rate law is dependent on the squared concentration of one reactant or,
- the rate law is dependent on the concentrations of two different reactants.
The basic rate laws for these two reaction types are, respectfully:
$$\text{rate}=k[A]^2$$
$$\text{rate}=k[A][B]$$
1. In the first case, the overall reaction can have more than one reactant. However, the reaction rate is found experimentally to actually depend only on the concentration of one of the reactants. This is typically the case when one of the reactants is in such excess that a change in its concentration is negligible. Here are some examples of this first type of second-order reaction:
$$\begin {align}&2NO_{2\,(g)} \xrightarrow {k} 2NO_{(g)} + O_{2\,(g)}\,\,;\text{rate}=k[NO_2]^2 \\&2HI_{(g)} \xrightarrow {k} H_{2\,(g)} + I_{2\,(g)} \,\,;\text{rate}=[HI]^2 \\&NO_{2\,(g)} + CO_{(g)} \xrightarrow {k} NO_{(g)} + CO_{2\,(g)}\,\,;\text{rate}=[NO_2]^2\end {align} $$
While the rate law may seem like it's following the coefficients for the unimolecular (one reactant) reactions, the rate law has actually been determined experimentally in each case.
2. In the second case, the rate is dependent on two reactants. The two reactants themselves are individually first-order (rate is dependent on that one reactant), but the overall reaction is considered second-order. The total order of a reaction is equal to the sum of the order of each reactant.
$$ \begin {align}&H^+_{(aq)} + OH^-_{(aq)} \xrightarrow {k} H_2O_{(l)}\,\,;\text{rate}=k[H^+][OH^-] \\&2NO_{2\,(g)} + F_{2\,(g)} \xrightarrow {k} 2NO_2F \,\,;\text{rate}=k[NO_2][F_2] \\&O_{3\,(g)} + Cl_{(g)} \xrightarrow {k} O_{2\,(g)} + ClO_{(g)}\,\,;\text{rate}=k[O_3][Cl]\end {align} $$
In this article, we will be covering both cases and look into how the reactant concentration can affect the rate.
Second-order Rate Law and Stoichiometry
While you may have noticed that some of the rate laws follow the stoichiometry, rate laws are actually experimentally determined.
Stoichiometry is the ratio of reactants to products in a chemical reaction.
Second-order Reaction Units
For each type of ordered reaction (zero-order, first-order, second-order, etc...), the rate constant, k. will have unique dimensional units depending on the overall order of the reaction. The reaction rate itself, however, will always be in the dimensions of M/s (molarity/second or moles/[second*liters]). This is because the rate of a reaction simply refers to the change in concentration over time. In the case of second-order reactions, the dimensions for the rate constant, k, are M-1 • s-1 or 1/[M • s]. Let's see why:
In what follows, we will square brackets, {...}, to contain the dimensional units. Thus, for a second-order reaction of the first type (rate is dependent on the squared concentration of one reactant), we will have:
$$rate\{ \frac{M}{s} \}=k\{ ? \}[A]^2\{ M^2 \}=k[A]^2\{ ? \} \{ M^2 \}$$
where, the bracket, {?}, represents the unknown dimension of the rate constant, k. Looking at the two brackets on the far right-hand side of the above equation we notice that the dimension of the rate constant has to be, {M-1 • s-1}, then:
$$rate\{ \frac{M}{s} \}=k\{ \frac{1}{M*s} \}[A]^2\{ M^2 \}=k[A]^2\{ \frac{1}{M*s} \} \{ M^2 \}=k[A]^2\{ \frac{M}{s} \}$$
Notice, now that giving the rate constant the correct dimensions, k{M-1 • s-1}, the formula for the rate law has the same dimensions on both sides of the equation.
Now, let's consider a second-order reaction of the second type (rate is dependent on the concentrations of two different reactants):
$$rate\{ \frac{M}{s} \}=k\{ ? \}[A]\{ M \}[B]\{ M \}=k[A][B]\{ ? \} \{ M^2 \}$$
where, the bracket, {?}, represents the unknown dimension of the rate constant, k. Again, looking at the two brackets on the far right-hand side of the above equation we notice that the dimension of the rate constant has to be, {M-1 • s-1}, then:
$$rate\{ \frac{M}{s} \}=k\{ \frac{1}{M*s} \}[A]\{ M \}[B]\{ M \}=k[A][B]\{ \frac{1}{M*s} \} \{ M \} \{ M \}=k[A][B]\{ \frac{M}{s} \}$$
Notice, again that giving the rate constant the correct dimensions, k{M-1 • s-1}, the formula for the rate law has the same dimensions on both sides of the equation.
The takeaway here is basically that, the units of the rate constant, k, are adjusted so that the rate law will always be in dimensions of molarity per second, M/s.
Second-order Reaction Formulas
If a given reaction has been determined to be second-order experimentally, we can use the integrated rate equation to calculate the rate constant based on the change in concentration. The integrated rate equation differs depending on which type of second-order reaction we are analyzing. Now, this derivation uses a lot of calculus, so we are just going to skip to the results (for those interested students please check out the "Deep dive" section below).
1. This equation is used for second-order reactions dependent on one reactant, the first type:
$$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$$
Where [A] is the concentration of reactant A at a given time, and [A]0 is the initial concentration of reactant A.
The reason why we set up the equation this way is for two reasons. The first is that it is now in linear form, y = mx+b, where; y = 1/[A], the variable, x = t, the slope is, m = k, and the y-intercept is, b = 1/[A0]. Based on the linear equation, we know that if the equation is graphed, k, will be the slope. The second reason is that the equation needs to be in the form of 1/[A], and not [A], because the equation is only linear this way. You'll see in a moment that if we graph the change in concentration over time, we will get a curve, not a line.
2. Now for the second type of second-order reaction. Note that if after the experimental determination of the rate law the reaction is found to be second-order and the concentrations of A and B are equal, we use the same equation as for type 1. If they aren't the same, the equation gets more complicated:
$$ln\frac{[A]}{[B]}=k([B]_0-[A]_0)t+ln\frac{[A]_0}{[B]_0}$$
where, [A] and [B], are the concentrations at time t, of A and B, respectively, and [A]0 and [B]0, are their initial concentrations. The key takeaway here is that when this equation is graphed, the slope is equal to, k([B]0-[A]0). Also, we need to take the natural log of the concentration to get a linear result.
For those of you who have taken calculus (or are just intrigued by it!), let's walk through the derivation of the rate law for the second-order reaction of the first type.
First, we set up our rate of change equation : $$-\frac{d[A]}{dt}=k[A]^2$$ This expression means that as the concentration of reactant, A, decreases with time, –d[A]/dt, it is equal to the given rate law, k[A]2.
Next, we rearrange the equation so both sides are in differential form, d(x). This is accomplished by multiplying both sides by dt: $$dt*-\frac{d[A]}{dt}=dt*k[A]^2$$ The two differentials, dt, on the left-hand side cancel: $$-{d[A]}=dt*k[A]^2$$ Now we multiply both sides by -1, and place the differential on the right-hand side at the end: $${d[A]}=-k[A]^2*dt$$ Then, we divide both sides by, [A]2, to get: $$\frac{d[A]}{[A]^2}=-kdt$$
Now that we have transformed the derivative into differentials, we can integrate. Since we are interested in the change in [A], over time, we integrate the rate law by starting with the expression on the left-hand side. We evaluate the definite integral from, [A] to [A]0, followed by integration of the expression on the right-hand side, from t to 0: $$\int_ {[A]_0}^{[A]} \frac{d[A]}{[A]^2}=\int_{0}^{t} -kdt$$ Let's first consider the integral on the left-hand side. To solve this integral, let's transform the variable [A] → x, then we have: $$\int_ {[A]_0}^{[A]} \frac{d[A]}{[A]^2}=\int_ {[A]_0}^{[A]} \frac{dx}{x^2}$$
Now we can evaluate the definite integral on the right-hand side, at the upper bound, [A], and lower bound, [A]0: $$\int_{[A]_0}^{[A]} \frac{dx}{x^2}=[\frac{-1}{x}]_{[A]_0}^{[A]}=\frac{-1}{[A]}-\frac{(-1)}{[A]_0}=\frac{-1}{[A]}+\frac{1}{[A]_0}$$ Now, let's go back and consider the integral on the right-hand side of the rate law:
$$\int _{0}^{t} -kdt=-k\int _{0}^{t} dt$$
To solve this integral, let's transform the differential dt → dx, then we have: $$-k\int _{0}^{t} dt=-k\int _{0}^{t} dx$$
Now evaluating the definite integral on the right-hand side, at the upper bound, t, and lower bound, 0, we get:
$$-k\int _{0}^{t} dx=-k[x]_{t}^{0}=-k*t-(-k*0)=-kt$$
Equating both sides of the results of the integration of the rate law, we get:
$$\frac{-1}{[A]}+\frac{1}{[A]_0}=-kt$$
or,
$$\frac{1}{[A]}- \frac{1}{[A]_0}=kt$$ Lastly, we rearrange this to get our final equation: $$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$$
Second-order Reaction Graphs
Let's first look at the graphs for the cases where the reaction is dependent only on one species.
The concentration of A over time decreases in an exponential or "curved" fashion. StudySmarter Original.
When we just graph the concentration over time, we get a curve like the one shown above. The graph only really helps us if we graph 1/[A] over time.
When the inverse of concentration over time is graphed, we see a linear relationship. StudySmarter Original.
As our equation suggests, the inverse of concentration over time is linear. We can use the equation of the line to calculate k and the concentration of A at a given time.
Given the equation of the line, what is the rate constant (k)? What is the concentration of A at 135 seconds? $$y=0.448+17.9$$
The first thing we need to do is compare this equation to the integrated rate equation:
$$\begin {align}&y=0.448x+17.9 \\&\frac{1}{[A]}=kt+\frac{1}{[A]_0}\end {align} $$
Comparing the equations, we see that the rate constant is, k = 0.448 M-1s-1. To get the concentration at 135 seconds, we just have to plug in that time for t and solve for [A].
$$\begin {align}&\frac{1}{[A]}=kt+\frac{1}{[A]_0} \\&\frac{1}{[A]}=0.448\frac{1}{M*s}(135\,s)+17.9\,M^{-1} \\&\frac{1}{[A]}=78.38\,M^{-1} \\&[A]=0.0128\,M\end {align} $$
We can also solve for k using the equation for slope when we are only given raw data.
At 5 seconds, the concentration of reactant A is 0.35 M. At 65 seconds, the concentration is 0.15 M. What is the rate constant?
To calculate k, we first need to change our concentration from [A] to 1/[A]. Then we can plug in the equation for slope. We must do this change since the equation is only linear in this form.
$$\begin {align}&\frac{1}{0.35\,M}=2.86\,M^{-1} \\&\frac{1}{0.15\,M}=6.67\,M^{-1} \\&\text{points}\,(5\,s,2.86\,M^{-1})\,(65\,s,6.67\,M^{-1}) \\&\text{slope}=\frac{y_2-y_1}{x_2-x_1} \\&\text{slope}=\frac{6.67\,M^{-1}-2.86\,M^{-1}}{65\,s-5\,s} \\&\text{slope}=k=0.0635\,M^{-1}s^{-1}\end {align} $$
Now for case 2: where the rate of reaction is dependent on two reactants A and B.
![Second-order reactions ln[A]/[B] over time StudySmarter](https://s3.eu-central-1.amazonaws.com/studysmarter-mediafiles/media/1865576/summary_images/lnA_B_over_time.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20240516%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20240516T112426Z&X-Amz-Expires=180000&X-Amz-SignedHeaders=host&X-Amz-Signature=238ad8574ee685c78f04131a3f9663e8766d62fb791405cc536eaab8c7766849)
When the change in ln[A]/[B] over time is graphed, we see a linear relationship. StudySmarter Original
Using this graph is a bit trickier than with type 1, but we can still use the equation of the line to calculate k.
Given the equation of the graph, what is the rate constant? [A]0 is 0.31 M
$$y=4.99x10^{-3}x-0.322$$
Like before, we need to compare the integrated rate equation to the linear equation
$$\begin {align}&y=4.99x10^{-3}x-0.322 \\&ln\frac{[A]}{[B]}=k([B]_0-[A]_0)t+ln\frac{[A]_0}{[B]_0} \\&k([B]_0-[A]_0)=4.99x10^{-3}\,s^{-1}\end {align}$$
We also have to use the y-intercept (ln[A]0/[B]0) to solve for [B]0 which we then can use to solve for k
$$\begin{align}&ln\frac{[A]_0}{[B_0}=-0.322 \\&\frac{[A]_0}{[B_0}=0.725 \\&[B]_0=\frac{[A]_0}{0.725} \\&[A]_0=0.31\,M \\&[B]_0=0.428\,M \\&k([B]_0-[A]_0)=4.99x10^{-3} s^{-1} \\&k(0.428\,M-0.31\,M)=4.99x10^{-3}s^{-1} \\&k=4.23x10^{-3}M^{-1}s^{-1}\end {align} $$
We can also use the equation to calculate the concentration of one of the reactants; however, we need to know the concentration of the other reactant at that time.
Half-life Formula for Second-order Reactions
There is a special form of the integrated rate equation we can use called the half-life equation.
A reactant's half-life is the time it takes for the concentration of the reactant to be halved. The basic equation is: $$[A]_{\frac{1}{2}}=\frac{1}{2}[A]_0$$
In this case, only second-order reactions that are dependent on one reactant have a half-life formula. For second-order reactions that are dependent on two reactants, the equation cannot be easily defined since A and B are different. Let's derive the formula:$$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$$$$[A]=\frac{1}{2}[A]_0$$$$\frac{1}{\frac{1}{2}[A]_0}=kt_{\frac{1}{2}}+\frac{1}{[A]_0} $$$$\frac{2}{[A_0}=kt_{\frac{1}{2}}+\frac{1}{[A]_0}$$$$\frac{1}{[A]_0}=kt_{\frac{1}{2}}$$$$t_{\frac{1}{2}}=\frac{1}{k[A]_0}$$
Now that we have our formula, let's work on a problem.
It takes 46 seconds for species A to decompose from 0.61 M to 0.305 M. What is k?
All we need to do is plug in our values and solve for k.
$$t_{\frac{1}{2}}=\frac{1}{k[A]_0}$$
$$46\,s=\frac{1}{k(0.61\,M)}$$$$k=\frac{1}{46\,s(0.61\,M)}$$$$k=0.0356\,\frac{1}{M*s}$$
Just remember that is only applicable for second-order reactions dependent on one species, not two.
Second Order Reactions - Key takeaways
- A second-order reaction is a reaction that's rate is dependent on either the squared concentration of one reactant or the concentrations of two reactants. The basic formulas for these two types are respectfully:$$\text{rate}=k[A]^2$$ $$\text{rate}=k[A][B]$$
The rate constant is in units of M-1s-1 (1/Ms)
The integrated rate equation for the first type of second-order reaction is: $$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$$
The integrated rate equation for the second type of second-order reaction is: $$ln\frac{[A]}{[B]}=k([B]_0-[A]_0)t+ln\frac{[A]_0}{[B]_0}$$
For the first case, the change in the inverse concentration over time is linear. For the second case, the change in the natural log of [A]/[B] over time is linear
A reactant's half-life is the time it takes for the concentration of the reactant to be halved.
The formula for half-life is \(t_{\frac{1}{2}}=\frac{1}{k[A]_0}\) . This is only applicable for the first type of second-order reaction
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