# 18.1: Rate of Reactions

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• define chemical kinetics and rates of reaction.
• write the rate expression and the units for the rate expression.
• define instantaneous rate.
• calculate instantaneous rate using a tangent line.

## Vocabulary

• chemical kinetics
• instantaneous rate
• rate of reaction

## Introduction

The focus of this chapter is chemical kinetics. Chemical kinetics is the study of chemical reactions rates and the factors that affect the rate of reactions. These factors include concentration, temperature, pressure, surface area, and the effect of a catalyst. For example, when food is placed in the refrigerator, the cold temperature keeps the food from decomposing by slowing the rate of reaction. Chemical kinetics plays an important role both in industry and in our daily lives. To begin, we will introduce some of the basic concepts of chemical kinetics.

## Change in Concentration Over Time

The term rate of reaction is used to denote the rate at which the products are formed in a time interval or the rate at which the reactants are consumed over a time interval. A reaction rate measures how fast or how slow a reaction is. The rusting of a piece of metal has a slow reaction rate because the iron oxidizes in the air over a relatively long time period. A forest fire has a fast reaction rate because it consumes trees in its path in a very short time interval. Reaction rates can be measured as the change in mass per unit time (grams/second) or the charge in molarity per unit time (mol/Ls\begin{align*}\mathrm{mol/L} \cdot \mathrm{s}\end{align*}).

Symbolically, the reaction rate is given the letter r\begin{align*}r\end{align*}. The reaction rate, then, can be written as follows:

r=concentrationtime\begin{align*}r = \frac {\triangle \text{concentration}} {\triangle \text{time}}\end{align*}

Remember that the symbol \begin{align*}\triangle\end{align*} means the “change in.”

Example:

For the reaction H2(g)+I2(g)2 HI(g)\begin{align*}\mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightarrow 2 \ \mathrm{HI}_{(g)}\end{align*}, under certain conditions, the [HI]=0.50 mol/L\begin{align*}[\mathrm{HI}] = 0.50 \ \mathrm{mol/L}\end{align*} at 25 s\begin{align*}25 \ \mathrm{s}\end{align*} and 0.75 mol/L\begin{align*}0.75 \ \mathrm{mol/L}\end{align*} at 40 s\begin{align*}40 \ \mathrm{s}\end{align*}. What is the rate of production of HI\begin{align*}\mathrm{HI}\end{align*}? Note: remember that the brackets indicate concentration.

Solution:

r=[HI]t=(0.750.50) mol/L(40.25) s=0.25 mol/L15 s=1.7×102 mol/Ls\begin{align*}r = \frac {\triangle [\mathrm{HI}]} {\triangle t} = \frac {(0.75 - 0.50) \ \text{mol/L}} {(40. - 25) \ \mathrm{s}} = \frac {0.25 \ \text{mol/L}} {15 \ \mathrm{s}} = 1.7 \times 10^{-2} \ \text{mol/L} \cdot \mathrm{s}\end{align*}

Therefore, the rate of production of HI\begin{align*}\mathrm{HI}\end{align*} is 1.7×102 mol/Ls\begin{align*}1.7 \times 10^{-2} \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}.

A blackboard discussion of reaction rate with an eye to developing the rate law and the equilibrium constant (8a) is available at http://www.youtube.com/watch?v=_HA1se_gyvs (7:20).

## Units for Rate of Reaction

Notice in the previous example that the units to measure the reaction rate are in mol/Ls\begin{align*}\mathrm{mol/L} \cdot \mathrm{s}\end{align*}. Therefore, the units are measuring the concentration/time or the M/time. These units allow for the comparison of rates. In other words, if all reaction rates were to use the same units, we can compare one rate to the other. For example, under a different set of conditions, the HI reaction was found to have a reaction rate of 2.5 mol/Ls\begin{align*}2.5 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}. We could then predict that the new set of conditions are favorable for this reaction since the reaction rate was faster for the production of HI in the same time interval.

## Graphing Instantaneous Rate

Instantaneous rate is defined as the rate of change at a particular moment. For example, a police officer stops a car for speeding. The radar gun on a police cruiser is set to measure the speed of a motorist as the motorist comes close to the cruiser. The driver of the vehicle is stopped doing 65 miles/hour\begin{align*}65 \ \mathrm{miles/hour}\end{align*} in a 50 miles/hour\begin{align*}50 \ \mathrm{miles/hour}\end{align*} zone. The cruiser measured the rate of speed at that instant in time when the driver passed the police officer. This is instantaneous rate. If we were to take all of the measures of instantaneous rate and graph them, we would obtain a curve of the overall speed (or the average speed) of the vehicle. The same is true for reactions. For reactions, the instantaneous rate is the rate of the reaction at a specific time in the reaction sequence. If you were to graph the rate of the reactant being consumed versus time, the graph would look like the figure below. As the reaction proceeds, the concentration of the reactants decreases over time.

The initial rate of the reaction is found at t=0 s\begin{align*}t = 0 \ \mathrm{s}\end{align*}, or when the reaction is just beginning. It is at this point when the maximum amount of the reactant is present. To find the instantaneous rate, a tangent line is drawn to this curve. The slope of this tangent line is then found. For example, say we wanted to know the instantaneous rate at t=2 s\begin{align*}t = 2 \ \mathrm{s}\end{align*}. After drawing the tangent line (see figure below), we can calculate the slope of the tangent line to find the instantaneous rate at t=2 s\begin{align*}t = 2 \ \mathrm{s}\end{align*}.

rate=y2y1x2x1=0.350.633.01.0=0.28 mol/L2.0 s=0.14 mol/L s\begin{align*}\text{rate} & = \frac {y_2 - y_1} {x_2 - x_1} = \frac {0.35 - 0.63} {3.0 - 1.0} = \frac {-0.28 \ \text{mol/L}} {2.0 \ \mathrm{s}} = -0.14 \ \text{mol/L} \cdot \ \mathrm{s}\\ \end{align*}

## Lesson Summary

• Chemical kinetics is the study of rates of chemical reactions and how factors affect rates of reactions. The term rate of reaction is used to denote the measure at which the products are formed over a time interval or the rate at which the reactants are consumed over a time interval.
• The units to measure the reaction rate are in mol/Ls\begin{align*}\mathrm{mol/L} \cdot \mathrm{s}\end{align*}.
• Instantaneous rate is defined as the rate of change at a particular moment.

The website below provides more details about measuring reaction rates.

## Review Questions

1. Given that the concentration of NO2(g)\begin{align*}\mathrm{NO}_{2(g)}\end{align*} is 0.40 mol/L\begin{align*}0.40 \ \mathrm{mol/L}\end{align*} at 45 s\begin{align*}45 \ \mathrm{s}\end{align*} and 0.85 mol/L\begin{align*}0.85 \ \mathrm{mol/L}\end{align*} at 80 s\begin{align*}80 \ \mathrm{s}\end{align*}, what is the rate of production of NO2(g)\begin{align*}\mathrm{NO}_{2(g)}\end{align*} in: NO2(g)+CO(g)NO(g)+CO2(g)\begin{align*}\mathrm{NO}_{2(g)} + \mathrm{CO}_{(g)} \rightarrow \mathrm{NO}_{(g)} + \mathrm{CO}_{2(g)}\end{align*}?
2. For the graph below, draw a tangent line at t=0.40 s\begin{align*}t = 0.40 \ \mathrm{s}\end{align*} and calculate the instantaneous rate.
3. Which expression represents the rate for the product formation for the reaction:Mg(s)+2 HCl(aq)MgCl2(aq)+H2(g)\begin{align*}\mathrm{Mg}_{(s)} + 2 \ \mathrm{HCl}_{(aq)} \rightarrow \mathrm{MgCl}_{2(aq)} + \mathrm{H}_{2(g)}\end{align*}?
1. rate=[Mg]t\begin{align*}\mathrm{rate} = \frac {\triangle [\mathrm{Mg}]} {\triangle \mathrm{t}}\end{align*}
2. rate=[HCl]t\begin{align*}\mathrm{rate} = \frac {\triangle [\mathrm{HCl}]} {\triangle \mathrm{t}}\end{align*}
3. rate=[MgCl2]t\begin{align*}\mathrm{rate} = \frac {\triangle [\mathrm{MgCl}_2]} {\triangle \mathrm{t}}\end{align*}
4. All of these are accurate representations of the rate.
4. Which statement represents a rate?
1. The speed of a car is 50 km/h\begin{align*}50 \ \mathrm{km/h}\end{align*}.
2. Half the product is produced.
3. A family consumes 5 L\begin{align*}5 \ \mathrm{L}\end{align*} of milk.
4. I ran for 45 minutes.\begin{align*}45 \ \mathrm{minutes.}\end{align*}
5. Which statement about the instantaneous rate of a reaction is correct?
1. The higher the rate, the smaller the slope of a line on a concentration-time graph.
2. The instantaneous rate is the slope of the tangent to a line on a concentration-time graph.
3. The instantaneous rate is the slope of the cosine to a line on a concentration-time graph.
4. All of these statements are correct.
6. What is the rate of production of NO\begin{align*}\mathrm{NO}\end{align*} gas if the concentration decreases from 0.32 mol/L\begin{align*}0.32 \ \mathrm{mol/L}\end{align*} at 56 s\begin{align*}56 \ \mathrm{s}\end{align*} and 0.94 mol/L\begin{align*}0.94 \ \mathrm{mol/L}\end{align*} at 78 s\begin{align*}78 \ \mathrm{s}\end{align*} for the reaction 4 NH3(g)+5 O2(g)4 NO(g)+6 H2O(g)\begin{align*}4 \ \mathrm{NH}_{3(g)} + 5 \ \mathrm{O}_{2(g)} \rightarrow 4 \ \mathrm{NO}_{(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(g)}\end{align*}?
1. 35 mol/Ls\begin{align*}-35 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
2. 2.8×102 mol/Ls\begin{align*}-2.8 \times 10^2 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
3. 2.8×102 mol/Ls\begin{align*}2.8 \times 10^{-2} \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
4. 35 mol/Ls\begin{align*}35 \ \mathrm{mol/L} \cdot \mathrm{s}\end{align*}
7. It takes 15 minutes\begin{align*}15 \ \mathrm{minutes}\end{align*} for the concentration of a reactant to decrease from 0.45 mol/L\begin{align*}0.45 \ \mathrm{mol/L}\end{align*} to 0.030 mol/L\begin{align*}0.030 \ \mathrm{mol/L}\end{align*}. What is the rate of reaction in \begin{align*}\mathrm{mol/L} \cdot \mathrm{s}?\end{align*}

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