Експериментальне визначення закону швидкості

Given an equation such as, $$\alpha A+\beta B \rightarrow \gamma C +\delta D$$ The rate of the reaction is given by, $$\text{rate}=k[A]^{a}[B]^{b}$$ where $a$ and $b$ have no relation to the balanced equation, and $k$ is a constant.

The most common way to determine $a$ and $b$ is the method of initial rates. In this method the reaction is performed several times and the concentrations of $A$ and $B$ are varied independently. The initial rate is measured each time, typically using a physical method such as absorption spectroscopy. A worked out, algebraic example is shown below.

Allow $r_{1}$ to be the initial reaction rate when the initial concentrations are $[A]=x$ and $[B]=y$. $$r_{1}=k\cdot x^{a}y^{b}$$ $r_{2}$ will be the initial reaction rate when the initial concentrations are $[A]=2x$ and $[B]=y$. $$r_{2}=k\cdot (2x)^{a}y^{b}=k\cdot 2^{a}x^{a}y^{b}$$ Finally, $r_{3}$ will be the initial reaction rate when the initial concentrations are $[A]=x$ and $[B]=2y$. $$r_{3}=k\cdot x^{a}(2y)^{b}=k\cdot x^{a}2^{b}y^{b}$$

In each of these three cases, $k$ has the same value. Thus, $$k=r_{1}x^{-a}y^{-b}=r_{2}\cdot 2^{-a}x^{-a}y^{-b}=r_{3}\cdot 2^{-b}x^{-a}y^{-b}$$ Therefore, $$\begin{align} \frac{r_{1}\cancel{x^{-a}}\cancel{y^{-b}}}{r_{2}\cancel{x^{-a}}\cancel{y^{-b}}}&=2^{-a} \\ log_{2}(\frac{r_{1}}{r_{2}})&=-a \\ a&=log_{2}(\frac{r_{2}}{r_{1}}) \end{align}$$ Likewise, $$\begin{align} \frac{r_{1}\cancel{x^{-a}}\cancel{y^{-b}}}{r_{3}\cancel{x^{-a}}\cancel{y^{-b}}}&=2^{-b} \\ log_{2}(\frac{r_{1}}{r_{3}})&=-b \\ b&=log_{2}(\frac{r_{3}}{r_{1}}) \end{align}$$ Thus, we can write our rate law as follows, $$\text{rate}=k[A]^{log_{2}(\frac{r_{2}}{r_{1}})}[B]^{log_{2}(\frac{r_{3}}{r_{1}})}$$

Now that $a$ and $b$ have been determined, $k$ can be found by using any of the three initial rates, $$r_{1}=kx^{log_{2}(\frac{r_{2}}{r_{1}})}y^{log_{2}(\frac{r_{3}}{r_{1}})}$$ $$r_{1}x^{log_{2}(\frac{r_{1}}{r_{2}})}y^{log_{2}(\frac{r_{1}}{r_{3}})}=k$$ Thus the rate may be written in the following manner, $$\text{rate}=r_{1}x^{log_{2}(\frac{r_{1}}{r_{2}})}y^{log_{2}(\frac{r_{1}}{r_{3}})}[A]^{log_{2}(\frac{r_{2}}{r_{1}})}[B]^{log_{2}(\frac{r_{3}}{r_{1}})}$$

In a chemical reaction, what happens at the microscopic scale? The atoms or molecules are bouncing around. Sometimes, they fly past one another. But, other times, they collide. What happens if they strike with the right amount of energy and the proper orientation? They react! The bonds break and reform into new configurations.

We've started studying chemical kinetics. It's a branch of physical chemistry that seeks to quantify and explain chemical reactions. Most often, we want to know how fast a reaction goes! Kinetics is all about determining the reaction rate. That's the change in concentration over time. So, it has units of molarity per second. We also connected stoichiometry and kinetics. We saw that the mole ratio influenced the reaction rate. Next, we looked at the reaction's order. The order tells us how changing a reactant's concentration alters the rate. Finally, we studied rate laws. A reaction's rate law is an equation that compares the rate versus the reactant's concentration.

So, we know all the individual components. But, how do we put it all together? How do we find a reaction's rate law? We have to do it experimentally.

How do we Find Reaction Orders?

We know how to write the rate laws for chemical reactions. But, how do we find the reaction order? We cannot determine the order just by looking at the balanced chemical equation. Instead, we have to find the order by experimentation. The experimental method is called the method of initial rates.

How do you perform the method of initial rates? The method's premise is to change the concentration of one reactant while keeping the others constant. You record the change in the reactant concentration over time. Your goal is to determine the initial rate of the reaction at time zero. So, you have a plot of reactant concentration versus time. If you know calculus, use the derivative at time zero to find the initial rate. If you don't know calculus, draw a tangent line through the origin. The initial rate is the tangent line's slope. You repeat this process for each reactant.

Example of an Initial Rates Table

You will have to solve problems based on the initial rates method. The question will have a table. The data it provides will be the trial, the reactants' molar concentrations, and the initial rates. Below, I've given you a generic table.

$$\begin{array}{c|c|c|c} Trial & [A](M) & [B](M) & Rate(\frac{[M]}{min}) \\ \hline 1 & 1.0[A] & 1.0[B] & R_{1} \\ 2 & 2.0[A] & 1.0[B] & R_{2} \\ 3 & 1.0[A] & 2.0[B] & R_{3} \end{array}$$

• Navigating the table.

• In trials one and three, A's concentration is the same. But, we have doubled B's concentration. So, we use trials one and three for finding B's reaction order.

• In trials one and two, A's concentration has doubled. We kept the B concentrations equal. So we would use trials one and two to calculate A's reaction order.

The Rate Ratio Equation

We need to find the reaction orders. To do that, we can take the ratios of two rate laws. By using the proportion, the rate constants cancel. We can also take advantage of the rules of exponents.

$\frac{Rate_{i}}{Rate_{f}}=\frac{k[A_{i}]^{X}[B_{i}]^{Y}}{k[A_{f}]^{X}[B_{f}]^{Y}} \rightarrow (\frac{[A]_{i}}{[A]_{f}})^{X} \times (\frac{[B]_{i}}{[B]_{f}})^{Y}$

Steps in Any Initial Rate Problem

For a generic reaction $A+B \rightarrow C$

• Use the rate ratio equation to solve for the reactant order.

• Use two trails that keep [A] the same. Solve for Y.

• Use two trials that keep [B] the same. Solve for X.

• Add the orders for A and B (X+Y) to find the overall order.

• Calculate the rate constant by plugging one trial's date into the rate law.

• Most rate questions will ask for the rate at a concentration of A and B. Plug the concentrations into the rate law and solve for the rate.

Generic Example Problem

We have a generic synthesis reaction, $A+B \rightarrow C$. Use the initial rates table to calculate the following.

a. The order for A and B.

b. The overall reaction order.

c. The rate constant.

d. The rate when [A]=0.33M and [B]=0.45M.

$$\begin{array}{c|c|c|c} Trial & [A](M) & [B](M) & Rate(\frac{[M]}{min}) \\ \hline 1 & 1.0 & 1.0 & 0.25 \\ 2 & 2.0 & 1.0 & 0.50 \\ 3 & 1.0 & 2.0 & 1.0 \end{array}$$

Use the rate ratio equation.

$\frac{Rate_{initial}}{Rate_{final}}=(\frac{[A]_{initial}}{[A]_{final}})^{X} \times (\frac{[B]_{initial}}{[B]_{final}})^{Y}$

Keep [A] the same. In this example, use trial 1 and trial 3.

$\frac{Rate_{trial 1}}{Rate_{Trial 3}}=(\frac{[A]_{trial 1}}{[A]_{trial 3}})^{X} \times (\frac{[B]_{trial 1}}{[B]_{trial 3}})^{Y}$

$\frac{.25}{1.0}=\frac{1.0}{1.0}^{X} \times \frac{1.0}{2.0}^{Y} \rightarrow 0.25= \frac{1.0}{2.0}^{Y} \rightarrow Y=\frac{log{(0.25)}}{log{(0.5)}} \rightarrow Y=2$

• One raised to any power is one. So, $(\frac{[A]_{trial 1}}{[A]_{trial 3}})^{X} = 1$.

• The reaction is second order for reactant B.

Keep [B] the same. In this example, use trials 1 and 2.

$\frac{Rate_{trial 1}}{Rate_{Trial 2}}=\frac{[A]_{trial 1}}{[A]_{trial 2}}^{X} \times \frac{[B]_{trial 1}}{[B]_{trial 2}}^{Y}$

$\frac{.25}{0.5}=\frac{1.0}{2.0}^{X} \times \frac{1.0}{1.0}^{Y} \rightarrow 0.5= \frac{1.0}{2.0}^{X} \rightarrow X=\frac{log{(0.5)}}{log{(0.5)}} \rightarrow X=1$

• One raised to any power is one. So, $(\frac{[B]_{trial 1}}{[B]_{trial 3}})^{Y} = 1$.

• The reaction is first order for reactant A.

Calculate the overall reaction order.

$overall\,order= X+Y= 1+2 =3$

The reaction is third order.

Calculate the rate constant. Now that we know the reaction order, we can calculate the rate constant using any trial.

$Rate_{trial1}=k\times[A_{1}]^{1}\times[B_{1}]^{2} \rightarrow .25\frac{M}{min}=k\times 1.0M \times 1.0M^{2} \rightarrow k=.25\frac{1}{M^{2}\times min}$

Calculate the rate when [A]=0.007M and [B]=0.003M.

$Rate=k[A]^{1}[B]^{2} \rightarrow rate=.25\frac{1}{M^{2}\times min}\times 0.33M \times (0.45M)^{2} \rightarrow rate=0.017\frac{M}{min}$

Specific Example Problem

Let's look at a more specific example. For this question, we'll examine the redox reaction between ammonium and nitrite ions. The products are nitrogen gas and water. Given the data table below calculate...

a. The order for each reactant.

b. The overall reaction order.

c. The rate constant.

d. The rate when $[NH^{+}_{4}]=0.07M$ and $[NO^{-}_{2}]=0.04M$.

$$\begin{array}{c|c|c|c} Trial & [NH^{+}_{4}](M) & [NO^{-}_{2}](M) & Rate(\frac{[M]}{sec}) \\ \hline 1 & .10 & .10 & .001 \\ 2 & .10 & .20 & .002 \\ 3 & .20 & .10 & .002 \end{array}$$

To solve part a, we need to use the ratio equation.

$\frac{Rate_{trial 1}}{Rate_{Trial 2}}=\frac{[A]_{trial 1}}{[A]_{trial 2}}^{X} \times \frac{[B]_{trial 1}}{[B]_{trial 2}}^{Y}$

We'll keep $[NH^{+}_{4}]$ the same and solve for Y. So we need trials 1 and 2 from our table.

$\frac{.001}{.002}=\frac{.10}{.10}^{X} \times \frac{.10}{.20}^{Y} \rightarrow 0.5=0.5^{Y} \rightarrow Y=1$

Now, we'll keep $[NO^{-}_{2}]$ the same and solve for X. So we need trials 1 and 3 from our table.

$\frac{.001}{.002}=\frac{.10}{.20}^{X} \times \frac{.10}{.10}^{Y} \rightarrow 0.5=0.5^{X} \rightarrow X=1$

So the reaction is first order with respect to both reactants. For part b, the overall order is second order $(X+Y=1+1=2)$.

To find the rate constant for part c, we insert one of the trials into our rate law. Let's use the first trial.

$rate=k[NH^{+}_{4}][NO^{-}_{2}] \rightarrow 0.001=K[.10][.10]\rightarrow k=0.1$

The trick is to find the units. Let's use dimensional analysis.

$\frac{M}{s}=k[M][M] \rightarrow k=\frac{M}{M^{2}\times s}\rightarrow k=\frac{0.1}{M\times s}$

For part d, we insert the values into our rate law.

$rate=\frac{0.1}{M\times s}[.07M][.04M]= 2.8\times 10^{-5} \frac{M}{s}$

As the scientific notation indicates, the rate is unbelievably small. That's because we had minuscule concentrations.

(Video) Practice Problem: Initial Rates and Rate Laws

by Professor Dave Explains

In this video, Professor Dave demonstrates an initial rates problem. He'll show you how to use experimental data to calculate the rate law. The problem has some complexity that's helpful for learning. You could see a similar problem on an exam. This is a great video to help you master the method of initial rates.