# 12.1 Chemical reaction rates  (Page 3/8)

 Page 3 / 8

## Relative rates of reaction

The rate of a reaction may be expressed in terms of the change in the amount of any reactant or product, and may be simply derived from the stoichiometry of the reaction. Consider the reaction represented by the following equation:

${\text{2NH}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)$

The stoichiometric factors derived from this equation may be used to relate reaction rates in the same manner that they are used to related reactant and product amounts. The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:

$-\phantom{\rule{0.2em}{0ex}}\frac{{\text{Δmol NH}}_{3}}{\text{Δ}t}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}{\text{2 mol}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\text{Δmol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}{\text{Δ}t}$

We can express this more simply without showing the stoichiometric factor’s units:

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δmol}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\text{Δmol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}{\text{Δ}t}$

Note that a negative sign has been added to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). If the reactants and products are present in the same solution, the molar amounts may be replaced by concentrations:

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{NH}}_{3}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δ}t}$

Similarly, the rate of formation of H 2 is three times the rate of formation of N 2 because three moles of H 2 form during the time required for the formation of one mole of N 2 :

$\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{H}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\text{Δ}\left[{\text{N}}_{\text{2}}\right]}{\text{Δ}t}$

[link] illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. We can see from the slopes of the tangents drawn at t = 500 seconds that the instantaneous rates of change in the concentrations of the reactants and products are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:

$\frac{2.91\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M\text{/s}}{9.71\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M\text{/s}}\phantom{\rule{0.2em}{0ex}}\approx 3$

## Expressions for relative reaction rates

The first step in the production of nitric acid is the combustion of ammonia:

$4{\text{NH}}_{3}\left(g\right)+5{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}4\text{NO}\left(g\right)+6{\text{H}}_{2}\text{O}\left(g\right)$

Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.

## Solution

Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are:

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{NH}}_{3}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{5}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{O}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[\text{NO}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{6}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{H}}_{2}\text{O}\right]}{\text{Δ}t}$

## Check your learning

The rate of formation of Br 2 is 6.0 $×$ 10 −6 mol/L/s in a reaction described by the following net ionic equation:

${\text{5Br}}^{\text{−}}+{\text{BrO}}_{3}{}^{\text{−}}+{\text{6H}}^{\text{+}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3Br}}_{2}+{\text{3H}}_{2}\text{O}$

Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{5}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{Br}}^{\text{−}}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{BrO}}_{3}{}^{\text{−}}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{H}}^{\text{+}}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{Br}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{H}}_{\text{2}}\text{O}\right]}{\text{Δ}t}$

## Reaction rate expressions for decomposition of h 2 O 2

The graph in [link] shows the rate of the decomposition of H 2 O 2 over time:

${\text{2H}}_{2}{\text{O}}_{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}_{2}\text{O}+{\text{O}}_{2}$

Based on these data, the instantaneous rate of decomposition of H 2 O 2 at t = 11.1 h is determined to be
3.20 $×$ 10 −2 mol/L/h, that is:

$-\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=3.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}{\text{mol L}}^{-1}{\text{h}}^{-1}$

What is the instantaneous rate of production of H 2 O and O 2 ?

## Solution

Using the stoichiometry of the reaction, we may determine that:

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{H}}_{2}\text{O}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\text{Δ}\left[{\text{O}}_{2}\right]}{\text{Δ}t}$

Therefore:

$\frac{1}{2}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}{\text{h}}^{-1}=\phantom{\rule{0.1em}{0ex}}\frac{\text{Δ}\left[{\text{O}}_{2}\right]}{\text{Δ}t}$

and

$\frac{\text{Δ}\left[{\text{O}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}{\text{h}}^{-1}$

## Check your learning

If the rate of decomposition of ammonia, NH 3 , at 1150 K is 2.10 $×$ 10 −6 mol/L/s, what is the rate of production of nitrogen and hydrogen?

1.05 $×$ 10 −6 mol/L/s, N 2 and 3.15 $×$ 10 −6 mol/L/s, H 2 .

## Key concepts and summary

The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.

## Key equations

• $\text{relative reaction rates for}\phantom{\rule{0.2em}{0ex}}a\text{A}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}b\text{B}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{a}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{b}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δ}t}$

## Chemistry end of chapter exercises

What is the difference between average rate, initial rate, and instantaneous rate?

The instantaneous rate is the rate of a reaction at any particular point in time, a period of time that is so short that the concentrations of reactants and products change by a negligible amount. The initial rate is the instantaneous rate of reaction as it starts (as product just begins to form). Average rate is the average of the instantaneous rates over a time period.

Ozone decomposes to oxygen according to the equation ${\text{2O}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3O}}_{2}\left(g\right).$ Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O 3 and the formation of oxygen.

In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction ${\text{Cl}}_{2}\left(g\right)+{\text{3F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2ClF}}_{3}\left(g\right).$ Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl 2 and F 2 and the formation of ClF 3 .

$\text{rate}=+\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{CIF}}_{3}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=-\frac{\text{Δ}\left[{\text{Cl}}_{2}\right]}{\text{Δ}t}\phantom{\rule{0.1em}{0ex}}=-\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}\left[{\text{F}}_{2}\right]}{\text{Δ}t}$

A study of the rate of dimerization of C 4 H 6 gave the data shown in the table:
${\text{2C}}_{4}{\text{H}}_{6}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{12}$

 Time (s) 0 1600 3200 4800 6200 [C 4 H 6 ] ( M ) 1.00 $×$ 10 −2 5.04 $×$ 10 −3 3.37 $×$ 10 −3 2.53 $×$ 10 −3 2.08 $×$ 10 −3

(a) Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.

(b) Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C 4 H 6 ]. What are the units of this rate?

(c) Determine the average rate of formation of C 8 H 12 at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).

A study of the rate of the reaction represented as $2A\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}B$ gave the following data:

 Time (s) 0 5 10 15 20 25 35 [ A ] ( M ) 1 0.952 0.625 0.465 0.37 0.308 0.23

(a) Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.

(b) Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [ A ]. What are the units of this rate?

(c) Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.

(a) average rate, 0 − 10 s = 0.0375 mol L −1 s −1 ; average rate, 12 − 18 s = 0.0225 mol L −1 s −1 ; (b) instantaneous rate, 15 s = 0.0500 mol L −1 s −1 ; (c) average rate for B formation = 0.0188 mol L −1 s −1 ; instantaneous rate for B formation = 0.0250 mol L −1 s −1

Consider the following reaction in aqueous solution:
${\text{5Br}}^{\text{−}}\left(\mathit{\text{aq}}\right)+{\text{BrO}}_{3}{}^{\text{−}}\left(\mathit{\text{aq}}\right)+{\text{6H}}^{\text{+}}\left(\mathit{\text{aq}}\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3Br}}_{2}\left(\mathit{\text{aq}}\right)+{\text{3H}}_{2}\text{O(}l\right)$

If the rate of disappearance of Br ( aq ) at a particular moment during the reaction is 3.5 $×$ 10 −4 M s −1 , what is the rate of appearance of Br 2 ( aq ) at that moment?

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