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By the end of this section, you will be able to:

- Define chemical reaction rate
- Derive rate expressions from the balanced equation for a given chemical reaction
- Calculate reaction rates from experimental data

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. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.

The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity.

For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. If we measure the concentration of hydrogen peroxide, H_{2}O_{2}, in an aqueous solution, we find that it changes slowly over time as the H_{2}O_{2} decomposes, according to the equation:

$${\text{2H}}_{2}{\text{O}}_{2}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{2H}}_{2}\text{O}(l)+{\text{O}}_{2}(g)$$

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

$$\begin{array}{cc}\hfill \text{rate of decomposition of}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{O}}_{2}& =-\phantom{\rule{0.2em}{0ex}}\frac{\text{change in concentration of reactant}}{\text{time interval}}\hfill \\ & =-\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{H}}_{2}{\text{O}}_{2}\right]}_{{t}_{2}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{\left[{\text{H}}_{2}{\text{O}}_{2}\right]}_{{t}_{1}}}{{t}_{2}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{t}_{1}}\hfill \\ & =-\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}\left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\text{\Delta}t}\hfill \end{array}$$

This mathematical representation of the change in species concentration over time is the rate expression for the reaction. 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 *t*_{1}; likewise,${[{\text{H}}_{2}{\text{O}}_{2}]}_{{t}_{2}}$ represents the molar concentration of hydrogen peroxide at a later time *t*_{2}; and Δ[H_{2}O_{2}] represents the change in molar concentration of hydrogen peroxide during the time interval Δ*t* (that is, *t*_{2}* − t_{1}). Since the reactant concentration decreases as the reaction proceeds, Δ[H_{2}O_{2}] is a negative quantity; we place a negative sign in front of the expression because reaction rates are, by convention, positive quantities. [link] provides an example of data collected during the decomposition of H_{2}O_{2}.*

To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:

$$\frac{-\text{\Delta}[{\text{H}}_{2}{\text{O}}_{2}]}{\text{\Delta}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{-(\text{0.500 mol/L}-\text{1.000 mol/L})}{(\text{6.00 h}-\text{0.00 h})}\phantom{\rule{0.1em}{0ex}}=\mathrm{0.0833\; mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{\mathrm{-1}}\phantom{\rule{0.2em}{0ex}}{\text{h}}^{\mathrm{-1}}$$

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:

$$\frac{-\text{\Delta}[{\text{H}}_{2}{\text{O}}_{2}]}{\text{\Delta}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{-(0.0625\phantom{\rule{0.2em}{0ex}}\text{mol/L}-0.125\phantom{\rule{0.2em}{0ex}}\text{mol/L})}{(24.00\phantom{\rule{0.2em}{0ex}}\text{h}-18.00\phantom{\rule{0.2em}{0ex}}\text{h})}\phantom{\rule{0.1em}{0ex}}=0.0104\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{\mathrm{-1}}\phantom{\rule{0.2em}{0ex}}{\text{h}}^{\mathrm{-1}}$$

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 (*t*_{0}). A few moments later, the instantaneous rate at a specific moment—call it *t*_{1}—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 (Δ*t*). 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 computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. If we plot the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H_{2}O_{2} at any time *t* is given by the slope of a straight line that is tangent to the curve at that time ([link]). We can use calculus to evaluating the slopes of such tangent lines, but the procedure for doing so is beyond the scope of this chapter.

Reaction Rates in Analysis: Test Strips for Urinalysis

Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine ([link]). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.

The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:

$${\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}+{\text{O}}_{2}\phantom{\rule{0.2em}{0ex}}\underset{\phantom{\rule{0.2em}{0ex}}\text{catalyst}\phantom{\rule{0.2em}{0ex}}}{\to}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{10}{\text{O}}_{6}+{\text{H}}_{2}{\text{O}}_{2}$$

$${\text{2H}}_{2}{\text{O}}_{2}+2{\text{I}}^{\text{\u2212}}\phantom{\rule{0.2em}{0ex}}\underset{\phantom{\rule{0.2em}{0ex}}\text{catalyst}\phantom{\rule{0.2em}{0ex}}}{\to}\phantom{\rule{0.2em}{0ex}}{\text{I}}_{\text{2}}+{\text{2H}}_{2}\text{O}+{\text{O}}_{2}$$

The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.

The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of *catalysis*, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a *false-negative* result). Waiting too long to assess the color change can lead to a *false positive* due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.

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:

$${\text{2NH}}_{3}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}(g)+{\text{3H}}_{2}(g)$$

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:

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

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

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

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:

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

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}:

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

[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:

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

Expressions for Relative Reaction Rates The first step in the production of nitric acid is the combustion of ammonia:

$$4{\text{NH}}_{3}(g)+5{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}4\text{NO}(g)+6{\text{H}}_{2}\text{O}(g)$$

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:

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

Check Your Learning
The rate of formation of Br_{2} is 6.0 $\times $ 10^{−6} mol/L/s in a reaction described by the following net ionic equation:

$${\text{5Br}}^{\text{\u2212}}+{\text{BrO}}_{3}{}^{\text{\u2212}}+{\text{6H}}^{\text{+}}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{3Br}}_{2}+{\text{3H}}_{2}\text{O}$$

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

Answer:

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{5}\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}[{\text{Br}}^{\text{\u2212}}]}{\text{\Delta}t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}[{\text{BrO}}_{3}{}^{\text{\u2212}}]}{\text{\Delta}t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}[{\text{H}}^{\text{+}}]}{\text{\Delta}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}[{\text{Br}}_{2}]}{\text{\Delta}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}[{\text{H}}_{\text{2}}\text{O}]}{\text{\Delta}t}$

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:

$${\text{2H}}_{2}{\text{O}}_{2}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{2H}}_{2}\text{O}+{\text{O}}_{2}$$

Based on these data, the instantaneous rate of decomposition of H_{2}O_{2} at *t* = 11.1 h is determined to be

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

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:

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

Therefore:

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

and

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

Check Your Learning
If the rate of decomposition of ammonia, NH_{3}, at 1150 K is 2.10 $\times $ 10^{−6} mol/L/s, what is the rate of production of nitrogen and hydrogen?

Answer:

1.05 $\times $ 10^{−6} mol/L/s, N_{2} and 3.15 $\times $ 10^{−6} mol/L/s, H_{2}.

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.

- $\text{relative reaction rates for}\phantom{\rule{0.2em}{0ex}}a\text{A}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}b\text{B}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{a}\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}[\text{A}]}{\text{\Delta}t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{b}\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}[\text{B}]}{\text{\Delta}t}$

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 ${\text{2O}}_{3}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{3O}}_{2}(g).$ 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 ${\text{Cl}}_{2}(g)+{\text{3F}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{2ClF}}_{3}(g).$ 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}.

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

A study of the rate of dimerization of C_{4}H_{6} gave the data shown in the table:

Time (s) | 0 | 1600 | 3200 | 4800 | 6200 |

[C_{4}H_{6}] (M) |
1.00 $\times $ 10^{−2} |
5.04 $\times $ 10^{−3} |
3.37 $\times $ 10^{−3} |
2.53 $\times $ 10^{−3} |
2.08 $\times $ 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 $2A\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}B$ gave the following data:

Time (s) | 0.0 | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 35.0 |

[A] (M) |
1.00 | 0.952 | 0.625 | 0.465 | 0.370 | 0.308 | 0.230 |

(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:

If the rate of disappearance of Br^{–}(*aq*) at a particular moment during the reaction is 3.5 $\times $ 10^{−4} *M* *s*^{−1}, what is the rate of appearance of Br_{2}(*aq*) at that moment?

- average rate
- rate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred

- initial rate
- instantaneous rate of a chemical reaction at
*t*= 0 s (immediately after the reaction has begun)

- instantaneous rate
- rate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time

- rate of reaction
- measure of the speed at which a chemical reaction takes place

- rate expression
- mathematical representation relating reaction rate to changes in amount, concentration, or pressure of reactant or product species per unit time

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