Chemistry Textbook

 

A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. In chemical reactions the amounts of products and reactants vary with time. Therefore, the rate of a chemical reaction can be measured by measuring changes in concentrations of reactants or products in a given amount of time (alternatively, changes in partial pressures can be measured if the reactants or products are gases, changes in light absorption can be measured if the reactants or products are colored, changes in conductivity can be measured if the reactants or products are electrolytes).

The rate of reaction is related to the change in the amount of a reactant or product per unit time. For example, the concentration of hydrogen peroxide, H₂O₂, in an aqueous solution changes slowly over time as it decomposes according to the equation:

The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:

In the above mathematical expression, the brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus, ${[{\text{H}}_2{\text{O}}_2]}_{t_1}$ represents the molar concentration of hydrogen peroxide at some time t₁; likewise,${[{\text{H}}_2{\text{O}}_2]}_{t_2}$ represents the molar concentration of hydrogen peroxide at a later time t₂; and Δ[H₂O₂] represents the change in molar concentration of hydrogen peroxide during the time interval Δt (that is, t₂ − t₁). Since the reactant concentration decreases as the reaction proceeds, Δ[H₂O₂] is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by −1. Figure 10.4 provides an example of data collected during the decomposition of H₂O₂ by measuring the concentration of hydrogen peroxide every 6 hours over the course of a day at a constant temperature of 40 °C.

The rate of decomposition in the first 6-hour period is calculated as:

Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:

This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t₀). A few moments later, the instantaneous rate at a specific moment—call it t₁—would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δt). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.

 

The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates over each short time intervals provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that which would yield similar results as in the case of short time interval measurements. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H₂O₂ at any time t is given by the slope of a straight line that is tangent to the curve at that time (Figure 10.5).

In the balanced equation for the decomposition of H₂O₂, 1 mol O₂ is formed for every 2 mol of H₂O₂ that reacts. Therefore, in a given interval of time, the change in moles of H₂O₂ can be related to the change in moles of O₂ as:

The above equation, essentially, means that O₂ appears half as fast as H₂O₂ disappears.

Similarly, from the molar ratios, the rate of the reaction will be equal to the rate of formation of O₂ (1 mol O₂ per mol reaction). The rate of the reaction will also be equal to half the rate of disappearance of H₂O₂ (2 mol H₂O₂ per mol reaction). This relationship can be expressed as:

For a general reaction:

The rate expression in terms of disappearance of reactants and appearance of products is given by:

$$\begin{array}{cc}& \begin{array}{|c|}\hline \mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=-\genfrac{}{}{0.1ex}{}{1}{\mathrm{a}}\genfrac{}{}{0.1ex}{}{\mathrm{\Delta }\left[\mathrm{A}\right]}{\mathrm{\Delta }\mathrm{t}}=-\genfrac{}{}{0.1ex}{}{1}{\mathrm{b}}\genfrac{}{}{0.1ex}{}{\mathrm{\Delta }\left[\mathrm{B}\right]}{\mathrm{\Delta }\mathrm{t}}=\genfrac{}{}{0.1ex}{}{1}{\mathrm{c}}\genfrac{}{}{0.1ex}{}{\mathrm{\Delta }\left[\mathrm{C}\right]}{\mathrm{\Delta }\mathrm{t}}=\genfrac{}{}{0.1ex}{}{1}{\mathrm{d}}\genfrac{}{}{0.1ex}{}{\mathrm{\Delta }\left[\mathrm{D}\right]}{\mathrm{\Delta }\mathrm{t}}\\ \hline\end{array}\hfill \end{array}$$

EXAMPLE 10.2

Reaction Rate Expressions for Decomposition of H₂O₂

For the decomposition of H₂O₂ over time:

$& 2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2\hfill & \hfill$

the instantaneous rate of decomposition of H₂O₂ at a certain time is determined to be
3.20 $\times$ 10⁻² mol/L/h. At the same time, (i) what is the instantaneous rate of production of O₂; and (ii) what is the rate of the reaction?

Solution:

The rate of decomposition of H₂O₂ is given:

$$-\frac{\text{Δ}[{\text{H}}_2{\text{O}}_2]}{\text{Δ}t}=3.20\times {10}^{\mathrm{-2}}{\text{mol L}}^{\mathrm{-1}}{\text{h}}^{\mathrm{-1}}$$

The expression for rate from reaction stoichiometry is:

$$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{ }-\frac{1}{2}\frac{∆\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}{∆t}=\mathrm{ }\frac{1}{2}\frac{∆\left[{\mathrm{H}}_2\mathrm{O}\right]}{∆t}=\mathrm{ }\frac{1}{2}\frac{∆\left[{\mathrm{O}}_2\right]}{∆t}$$  

(i) We need to calculate rate of production of O2 (ie. Δ[O2]/Δt).

     Isolating parts of the rate expression that are relevant to H2O2 and O2:

$$\frac{1}{2}\mathrm{ }\left(-\frac{∆\left[{{\mathrm{H}}_2\mathrm{O}}_2\right]}{∆\mathrm{t}}\right)= \frac{∆\left[{\mathrm{O}}_2\right]}{∆\mathrm{t}}\mathrm{ }\mathrm{ }$$

$$\frac{1}{2}\times 3.20\times {10}^{\mathrm{-2}}\text{mol}{\text{L}}^{\mathrm{-1}}{\text{h}}^{\mathrm{-1}}=\frac{\text{Δ}[{\text{O}}_2]}{\text{Δ}t}$$

$$\frac{\text{Δ}[{\text{O}}_2]}{\text{Δ}t}=1.60\times {10}^{\mathrm{-2}}\text{mol}{\text{L}}^{\mathrm{-1}}{\text{h}}^{\mathrm{-1}}$$

     Another way to solve:

$$\frac{3.2\times {10}^{-2} mol L^{-1}{h}^{-1} H_2{O}_2 }{}. \frac{1 mol O_2}{2 mol H_2{O}_2} = 1.6\times {10}^{-2} mol L^{-1}{h}^{-1} O_2$$

(ii) The rate of the reaction is given as:

$\begin{array}{ccc}& \mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=-\genfrac{}{}{0.1ex}{}{1}{2}\genfrac{}{}{0.1ex}{}{\mathrm{\Delta }\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]}{\mathrm{\Delta }\mathrm{t}}=\genfrac{}{}{0.1ex}{}{\mathrm{\Delta }\left[{\mathrm{O}}_2\right]}{\mathrm{\Delta }\mathrm{t}}=1.6\times 10^{-2}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\mathrm{h}{\mathrm{r}}^{-1}\hfill & \hfill \end{array}$

     Another way to solve:

$$\frac{3.2\times {10}^{-2} mol L^{-1}{h}^{-1} H_2{O}_2 }{}. \frac{1 mol reaction}{2 mol H_2{O}_2} = 1.6\times {10}^{-2} mol L^{-1}{h}^{-1}$$  

Check Your Learning: If the rate of decomposition of ammonia, NH₃, at 1150 K is 2.10 $\times$ 10⁻⁶ mol/L/s, (i) what is the rate of production of nitrogen? (ii) what is the rate of production of hydrogen? (iii) what is the rate of the reaction?

$\begin{array}{ccc}& 2\mathrm{N}{\mathrm{H}}_3\to {\mathrm{N}}_2+3{\mathrm{H}}_2\hfill & \hfill \end{array}$

Answer:

(i) 1.05 $\times$ 10⁻⁶ mol/L/s N₂; (ii) 3.15 $\times$ 10⁻⁶ mol/L/s H₂; (iii) rate = 1.05 $\times$ 10⁻⁶ mol/L/s.