Стехіометрія та швидкість реакції

I have an older house. It got built in the 1970s. So, it's about 50 years old! In the 1970s, they used copper pipes for all the plumbing. Copper is a transition metal with very low reactivity to water. However, water treatment plants chlorinate the water. Chlorine reacts with water to form hypochlorous acid. That's great for killing microbes. But, it reacts with copper. Hypochlorous acid is a strong oxidizing agent. So, it oxidizes copper metal into copper ions. The ions are soluble in water. So, the pipe degrades over time. That's why most plumbing today is polyvinyl chloride (PVC). PVC is resistant to water, acids, and bases. In theory, it will never corrode.

Copper pipes are just one way that we use kinetics outside the lab. We are constantly thinking about how long something lasts. But, we need to know how to apply kinetics to chemical reactions. Today, we'll combine kinetics and stoichiometry. By combining the two, we can find how quickly the reactants deplete and the products form.

Rates and Stoichiometry

We can combine reaction rates with our knowledge of stoichiometry.

Let's say that we have a generic reaction $A \rightarrow B + C$. For every mole of A consumed, we produce one mole of B and C. What about the rates? If A depletes at a specific rate, B and C form at that same rate. We can write this mathematically as, $-\frac{\Delta[A]}{\Delta\,t}=\frac{\Delta[B]}{\Delta\,t}=\frac{\Delta[C]}{\Delta\,t}$. We have the molar concentration (moles per liter) divided by time. The first term is negative because the reaction consumes A.

What happens when our equation looks like, $aA \rightarrow bB +cC$? The lower case letters denote the coefficients. In this case, we use the mole ratios.

$-\frac{\Delta[A]}{\Delta\,t}\times \frac{b}{a}=\frac{\Delta[B]}{\Delta\,t}$

$-\frac{\Delta[A]}{\Delta\,t}\times \frac{c}{a}=\frac{\Delta[C]}{\Delta\,t}$.

Are you wondering if the units work correctly? In the second example problem, I included the dimensional analysis. You can see that they do work correctly!

Steps in a Rate-Stoichiometry Problem

• Step 1: Write the chemical equation.

• Step 2: Balance the chemical equation.

• Step 3: Use the data to calculate the rate. If you already know it, move to step four.

• Step 4: Calculate the other rates using the mole ratios.

• Step 5: Answer the question if it asks for something other than the rates.

Finding Rates from Tables

Sometimes, your teacher will give you a table of data. Typically, it will have the reactant's concentration versus time. You can use the table to find the average rate. Let's look at an example.

Again, let's say we have the generic reaction, $aA \rightarrow bB +cC$. The data for the consumption of A is in the table.

$$\begin{array}{r|l} Time (s) & [A] (M) \\ \hline 0 & 1.00 \\ 10 & 0.94 \\ 20 & 0.79 \\ 30 & 0.57 \\ 40 & 0.22 \\ 50 & 0.02 \end{array}$$

How would we find the average rate from 0 seconds to 30 seconds?

We would use the formula, $avarge\,rate =\frac{[A]_{final}-[A]_{initial}}{t_{final}-t_{initial}}$.

What do we get if we plug in the values at 0 seconds and 30 seconds?

$avarge\,rate =\frac{0.57M-1.00M}{30.0s-0.0s}=\frac{-0.43M}{30.0s}=-0.014\frac{M}{s}$.

The answer is negative because we deplete A. You can use this procedure to find the average rate over any given time.

Example Problems

1. The table below gives the rate data for the reaction of the organic compound 1-chlorobutane with liquid water. The product is butanol. Find the average rate of 1-chlorobutane consumption from 100 to 400 seconds.

$$\begin{array}{r|l} Time (s) & [C_{4}H_{9}Cl] (M)\\ \hline 0 & 1.00 \\ 100 & 0.87 \\ 200 & 0.62 \\ 300 & 0.51 \\ 400 & 0.39 \\ 500 & 0.32 \end{array}$$

$Average\,rate=\frac{\Delta [C_{4}H_{9}Cl]}{\Delta t}=\frac{[C_{4}H_{9}Cl]_{final}-[C_{4}H_{9}Cl]_{initial}}{t_{final}-t_{initial}}= \frac{0.39M-0.87M}{400s-100s}=\frac{-0.48M}{300s}=-0.0016\frac{M}{s}$

1. The decomposition reaction of dinitrogen pentoxide produces nitrogen dioxide and oxygen. All three are in the gas phase. Nitrogen pentoxide's average decomposition rate is $1.6\times10^{-3} \frac{M}{s}$. What are the average formation rates for nitrogen dioxide and oxygen?

• Step 1: Write the chemical equation.

$N_{2}O_{5(g)} \rightarrow NO_{2(g)} + O_{2(g)}$

• Step 2: Balance the chemical equation.

$2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$

• Step 3: Use the mole ratios to calculate the rates.

$rate_{NO_{2(g)}}=rate_{N_{2}O_{5(g)}} \frac{\frac{mol_{N_{2}O_{5(g)}}}{L}}{s}\times\frac{4 mols_{NO_{2(g)}}} {2mols_{N_{2}O_{5(g)}}} \rightarrow 1.6\times10^{-3}\frac{\frac{mol_{N_{2}O_{5(g)}}}{L}}{s}\times\frac{4 mols_{NO_{2(g)}}}{2 mols_{N_{2}O_{5(g)}}}=0.0032\frac{M}{s}$

$rate_{O_{2(g)}}=rate_{N_{2}O_{5(g)}} \frac{\frac{mol_{N_{2}O_{5(g)}}}{L}}{s}\times\frac{1 mols_{O_{2(g)}}} {2mols_{N_{2}O_{5(g)}}} \rightarrow 1.6\times10^{-3}\frac{\frac{mol_{N_{2}O_{5(g)}}}{L}}{s}\times\frac{1 mols_{O_{2(g)}}}{2 mols_{N_{2}O_{5(g)}}}=0.0008\frac{M}{s}$

(Video) Relative Rates from Stoichiometric Coefficients

by Denovo Tutor

(Video) Relative Rates | Chemistry Problem

by Denovo Tutor

In these two short videos, the Denovo Tutor explains relative rates and their connection to the mole ratios. This first video shows you an easy method for solving problems. The second video shows you an example problem.