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 relate reactant and product amounts. The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:

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

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

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

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:

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

Similarly, the rate of formation of H₂ is three times the rate of formation of N₂ because three moles of H₂ form during the time required for the formation of one mole of N₂:

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

The figure below 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:

\(\cfrac{2.91\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M\text{/s}}{9.70\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M\text{/s}}\phantom{\rule{0.2em}{0ex}}\approx 3\)

A graph is shown with the label, “Time ( s ),” appearing on the x-axis and, “Concentration ( M ),” on the y-axis. The x-axis markings begin at 0 and end at 2000. The markings are labeled at intervals of 500. The y-axis begins at 0 and includes markings every 1.0 times 10 superscript negative 3, up to 4.0 times 10 superscript negative 3. A decreasing, concave up, non-linear curve is shown, which begins at about 2.8 times 10 superscript negative 3 on the y-axis and nearly reaches a value of 0 at the far right of the graph at the 2000 marking on the x-axis. This curve is labeled, “[ N H subscript 3].” Two additional curves that are increasing and concave down are shown, both beginning at the origin. The lower of these two curves is labeled, “[ N subscript 2 ].” It reaches a value of approximately 1.25 times 10 superscript negative 3 at 2000 seconds. The final curve is labeled, “[ H subscript 2 ].” It reaches a value of about 3.9 times 10 superscript negative 3 at 2000 seconds. A red tangent line segment is drawn to each of the curves on the graph at 500 seconds. At 500 seconds on the x-axis, a vertical dashed line is shown. Next to the [ N H subscript 3] graph appears the equation “negative capital delta [ N H subscript 3 ] over capital delta t = negative slope = 1.94 times 10 superscript negative 6 M / s.” Next to the [ N subscript 2] graph appears the equation “negative capital delta [ N subscript 2 ] over capital delta t = negative slope = 9.70 times 10 superscript negative 7 M / s.” Next to the [ H subscript 2 ] graph appears the equation “negative capital delta [ H subscript 2 ] over capital delta t = negative slope = 2.91 times 10 superscript negative 6 M / s.”

This graph shows the changes in concentrations of the reactants and products during the reaction \(2{\text{NH}}_{3}\text{}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}+{\text{3H}}_{2}.\) The rates of change of the three concentrations are related by their stoichiometric factors, as shown by the different slopes of the tangents at t = 500 s.

Example

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:

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

Example

Reaction Rate Expressions for Decomposition of H₂O₂

A graph is shown with the label, “Time ( h ),” appearing on the x-axis and “[ H subscript 2 O subscript 2 ] ( mol L superscript negative 1)” on the y-axis. The x-axis markings begin at 0 and end at 24. The markings are labeled at intervals of 6. The y-axis begins at 0 and includes markings every 0.200, up to 1.000. A decreasing, concave up, non-linear curve is shown, which begins at 1.000 on the y-axis and nearly reaches a value of 0 at the far right of the graph around 10 on the x-axis. A red tangent line segment is drawn on the graph at the point where the graph intersects the y-axis. A second red tangent line segment is drawn near the middle of the curve. A vertical dashed line segment extends from the left endpoint of the line segment downward to intersect with a similar horizontal line segment drawn from the right endpoint of the line segment, forming a right triangle beneath the curve. The vertical leg of the triangle is labeled “capital delta [ H subscript 2 O subscript 2 ]” and the horizontal leg is labeled, “capital delta t.”

This graph shows a plot of concentration versus time for a 1.000 M solution of H₂O₂. The rate at any instant is equal to the opposite of the slope of a line tangential to this curve at that time. Tangents are shown at t = 0 h (“initial rate”) and at t = 10 h (“instantaneous rate” at that particular time).

The graph in the figure above shows the rate of the decomposition of H₂O₂ 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 H₂O₂ at t = 11.1 h is determined to be

3.20 \(×\) 10⁻² mol/L/h, that is:

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

What is the instantaneous rate of production of H₂O and O₂?

Solution

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

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

Therefore:

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

and

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

Relative Rates of Reaction