Introduction
Chemical kinetics involves analyzing and determining the rates at which reactions occur. Through analyzing experimental data, we can hypothesize a reaction mechanism that shows the stepwise process as bonds are broken and formed during the reaction. Reactions present in nature and life occur at widely different rates. Catalysts speed up the rates of chemical reactions. Biological catalysts, or enzymes are crucial to life, without which many reactions would proceed too slowly.
Reaction Rates
Reaction rates are typically expressed in units of molarity per second (M/s). This is a change in concentration over a time interval. Let's first consider the average rate of a reaction. Let's consider the hypothetical reaction as follows:
\(\ce{A -> B}\)
How can we express the rate at which B, the product, is produced?
\(\text{Rate Production } B = \frac{[B]_{t2} - [B]_{t1}}{t2 - t1} = \frac{\Delta[B]}{\Delta t}\)
Here, we have the change in concentration of B over the change in time. By convention, a rate is always expressed as positive in chemistry. Let's consider the perspective of the rate from the reactant, A. In this case:
\(\text{Rate Disappearance } A = -\frac{[A]_{2} - [A]_{1}}{t_{2} - t_{1}} = -\frac{\Delta[A]}{\Delta t}\)
Since the concentration of A is decreasing as the reaction proceeds, [A]2 will be less than [A]1, and ΔA would be negative. The negative sign switches the sign, making the rate positive.
Given this, let's consider the following example: At 30 seconds the concentration of A is 0.8 M, and at 80 seconds it is 1.2 M. What is the average rate of this reaction?
\(\text{Average Rate} = -\frac{\Delta[A]}{\Delta t} = -\frac{0.8 \, \text{M} - 1.2 \, \text{M}}{80 \, \text{s} - 30 \, \text{s}} = -\left(-\frac{0.4 \, \text{M}}{50 \, \text{s}}\right) = 0.008 \, \text{M/s}\)
Let's say the stoichiometry between reacants and products is not 1:1 as in the previous example. Let's consider the reaction between nitrogen monoxide and hydrogen gas to form nitrogen gas and water vapor.
\(\ce{2NO(g) + 2H2(g) -> N2(g) + 2H2O(g)} \)In calculus, the derivative demonstrates the sensitivity of an function's output relative to its input. Taking the derivative can give us the instantaneous rate of change. Consider the classical example from mechanics with the following position vs time graph:
Here, the position vs. time is described by the function x(t) = t2. If we want to know the instantaneous velocity at any point along our graph, we take the derivative: x'(t) = 2t. This is analagous to finding the slope of a tangent line at any given point along our curve. For example, at t = 5 s, the velocity is x'(5) = 2(5) = 10 m/s, which is the same as the slope of our tangent line (y = 10x - 25).
A similar analysis can be performed in chemical kinetics. Consider the following curve below which shows a concentration vs time graph for the rate of decomposition of N2O5 into NO2 and O2:
We can similarly take the derivative at some point along this curve to determine the instantaneous rate. As we will soon see, the curve and underlying relationship that results from observing the concentration vs. time for a given chemical reaction tells us something about how the rate depends on reactant concentrations, something that can be described using a rate law.
Rate Laws
A rate law describes the rate of a chemical reaction based on the concentrations of the reactants. We will consider zeroth order, first order, and second order reactions.
Let's consider the following reaction:
\(\ce{NH4+(aq) + NO2-(aq) -> N2(g) + 2H2O(l)}\)
Here, we have the reaction and ammonium and nitrogen dioxide to form nitrogen gas and water. Experimentally, one can conduct a series of experiments in which the concentration of one of the reacants is held constant while the other is changed and the rate is measured. Consider the following data for a series of experiments:
| Experiment Number | Initial NH4+ Concentration (M) | Initial NO2- Concentration (M) | Observed Initial Rate (M/s) |
|---|---|---|---|
| 1 | 0.0100 | 0.200 | 5.4x10-7 |
| 2 | 0.0200 | 0.200 | 10.8x10-7 |
| 3 | 0.0400 | 0.200 | 21.5x10-7 |
| 4 | 0.200 | 0.0202 | 10.8x10-7 |
| 5 | 0.200 | 0.0404 | 21.6x10-7 |
| 6 | 0.200 | 0.0808 | 43.3x10-7 |
Again, consider the general form of a rate law:
\(\text{Rate} = k[\text{A}]^w[\text{B}]^x[\text{C}]^y...\)
The exponents associated with each of the reactant concentrations (w, x, y) are known as the reaction orders. The sum of the orders with respect to each rectant is known as the overall reaction order. The epxonents give us information about how the rate of the reaction changes in response to changing reactant concentration. For example, consider the following rate law:
\(\text{Rate} = k[\text{A}]^2[\text{B}]^3[C]\)
We would say that this reaction is second order with respect to A, third order with respect to B, and first order with respect to C. Summing these exponents we get: 2 + 3 + 1 = 6, so we would say the reaction is sixth order overall. Again, we can get valuable information from the exponents. What would be the effect of doubling [C] on the rate of the reaction? This would result in an 8x increase in reaction rate, as [2]3 = 8.
Again, while the exponents may at times be the same as coefficients in the balanced chemical equation, this is not always the case. The rate law must be determined experimentally. Generally, the reactions orders are 0, 1, and 2. In the case of zero order with respect to some reactant, changing its concentration has no effect on the overall rate, as [A]0 = 1. While the rate of the reaction overall is concentration dependent, the rate constant, k, is concentration indepedent, and will be discussed in more detail shortly.
Reaction Order
The rate laws we have been looking at up to this point have been First Order
Consider a reaction in which the rate depends on the concentration of a single reactant, and the rate is first order with respect to this reactant:
\(\text{Rate} = \frac{-\Delta[\text{A}]}{\Delta t} = k[\text{A}]
\)
This is a differential equation that shows the change in the concentration of A with respect to time. How could we represent this as an equation that releates the initial concentration to the concentration at some other time? Using calculus we can express our differential rate law (which shows the how the rate of the reaction changes with respect to changing reactant concentrations) as:
\(\frac{-d[\text{A}]}{dt} = k[\text{A}]
\)
This demonstrates the dependence of rate on concentration. What if we integrate? Here, we use separation of variables.
\begin{align*}
\text{-d}[A]/\text{d}t &= k[A] \\
\text{d}[A] &= -k[A]\text{d}t \\
\frac{1}{[A]}\text{d}[A] &= -k\text{d}t \\
\int_{[A]_0}^{[A]_t} \frac{1}{[A]}\text{d}[A] &= \int_{0}^{t} -k\text{d}t \\
\ln[A] \bigg|_{[A]_0}^{[A]_t} &= -kt \bigg|_0^t \\
\ln[A]_t - \ln[A]_0 &= -kt \\
\ln[A]_t &= \ln[A]_0 - kt \\
\end{align*}
To review, we have two variables here, [A] and t. First, we get [A] on one side of the equation and then t on the other side of the equation. We are interested in the initial concentration and the concentration at some later time t, so we integrate both sides accordingly. We are left with the integrated rate law for a first order reaction:
\begin{align*}
\ln[A]_t = \ln[A]_0 - kt \\
\end{align*}
The integrated rate law is useful for solving a number of problems. For example, let's say we want to know the concentration of [A] at some later time t. We could algebraically rearrange the integrated rate law to solve for [A]t as follows:
\begin{align*}
\ln[A]_t &= \ln[A]_0 - kt \\
e^{\ln[A]_t} &= e^{\ln[A]_0 - kt} \\
[A]_t &= \frac{e^{\ln[A]_0}}{e^{kt}} \\
[A]_t &= \frac{[A]_0}{e^{kt}} \\
[A]_t &= [A]_0e^{-kt}
\end{align*}
Now, we have an equation to solve to solve for the concentration of [A] at some later point t based on the initial concentration [A]0. What if we were to graph the concentration vs. time and the ln concentration vs time for a first order reaction?
Consider the instantaneous rate of change at different points along the graph for the [N2O5] vs. time. The instantaneous rate of change starts out steeply before continuing to decrease as the [N2O5] decreases, confirming the fact that the rate of decrease in concentration is directly proportional to the concentration.
Again, we saw this reaction earlier (the decomposition of N2O5 to NO2 and O2, a first order process). What happens when we plot the natural log of the concentration of N2O5 vs. time? Click the button above to toggle between the two charts. Taking the ln results in a linear graph, or straight-line plot, which can be described by the integrated rate law we derived earlier:
\begin{align*}
\ln[A]_t = \ln[A]_0 - kt \\
\end{align*}
Here, the slope of the line for ln concentration vs. time is equal to -k. If the plot of the ln concentration vs. time graph for a reaction is linear, that reaction is a first order process. In other words, the nature of the graphs can be used to determine the reaction order of an unknown reaction. We will see this again for zeroth order and second order reactions, for which we will go through a similar process as we did here for first order recations.
| Experiment | Initial [A] | Observed Initial Rate (M/s) |
|---|---|---|
| 1 | 0.1 | 1 |
| 2 | 0.2 | 4 |
| 3 | 0.4 | 16 |
In a zero-order reaction, the rate is independent of the reactant concentration, and is therefore always equal to the rate constant k. In this process, a fixed amount of reactant reacts per unit time, regardless of concentration. The differential rate law is as follows:
\(\text{Rate} = -\frac{\Delta [A]}{\Delta t} = k\)
Again, by separation of variables we can find the integrate rate law:
\begin{align*}
-\frac{d[A]}{dt} &= k \\
d[A] &= -k dt \\
\int_{[A]_0}^{[A]_t} d[A] &= \int_{0}^{t} -k dt \\
[A] \Big|_{[A]_0}^{[A]_t} &= -kt \Big|_{0}^{t} \\
[A]_t - [A]_0 &= -kt \\
[A]_t &= [A]_0 - kt
\end{align*}
Graphing the integrated rate law we find:
Both the differential rate law (concentration vs. time) and integrated rate law graphed as concentration vs. time are linear. The slope in the graph represented by the integrated rate law is -k.
| Zeroth Order | First Order | Second Order | |
|---|---|---|---|
| Differential Rate Law | Rate = \( - \frac{\Delta[A]}{\Delta t} = k \) | Rate = \( - \frac{\Delta[A]}{\Delta t} = k[A] \) | Rate = \( - \frac{\Delta[A]}{\Delta t} = k[A]^2 \) |
| Integrated Rate Law | \([A] = [A]_0 - kt\) | \(\ln[A] = \ln[A]_0 - kt\) | \(\frac{1}{[A]} = \frac{1}{[A]_0} + kt\) |
| Half-Life | \( t_{1/2} = \frac{[A]_0}{2k} \) | \( t_{1/2} = \frac{0.693}{k} \) | \( t_{1/2} = \frac{1}{k[A]_0} \) |
| Units of \( k \) | M/s | 1/s | \( M^{-1}s^{-1} \) |
Other Factors Affecting Rate
A quick note: you may have encountered the equilibrium constant, K, before. The equilibrium constant is the ratio of the product concentrations to reactant concentrations for a reversible reaction at equilibrium, and is calculated by taking the concentrations raised to the power of their stoichiometric coefficients. We have been discussing k, the rate constant. We will now look at the affects of temperature and catalysts on the rate of chemical reactions.
The rate constant, k, is a temperature dependent factor. As the temperature increases, the rate constant increases as well. This is described empirically by the Arrhenius equation:
\[ k = A \times e^{\frac{{-E_a}}{{RT}}} \] Where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. The rate constant is directly proportional to temperature.Catalysts increase the rate of chemical reactions without being consumed in the process. Catalysts can be homogenous: existing in the same phase as the reacants, heterogeneous: existing in a different phase than the reactants, or biological. Biological catalysts are known as enzymes. Catalysts works by lowering the activation energy (Ea) of a reaction.