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Rates of Reaction

>> Parts of this equation/concept include:

A -	Determining Rates >> Average Rates, Equation 14.6 >> Instantaneous Rates >> Initial Rates
B -	Stoichiometric Relationships Between Rates

>> Average Rates, Equation 14.6

Average rates can be determined from any two initial and final time concentration values using Equation 14.6.

rate

=

[X] t

=

([X]f – [X]i) (tf – ti)

(Equation 14.6)

Positive and negative values of rate will come about naturally. (Reactants decrease in concentration; products increase.) Like with thermodynamic measurements (see chapters 11 and 13), the sign indicates direction.

>> Example 1

What is the average rate based on the following data?

1. The concentration of reactant changes from 0.50 M to 0.01 M in 1.5 minutes.
2. The concentration of product is 0.056 M after 75 seconds.
3. The [X]₀ = 1.00 M at t = 0 and [X] = 0.34 M at t = 50 s.

Solution:

The average rates are determined from initial and final concentration and time data. The tradition is to subtract final from initial value.

1. Initial concentration = 0.50 M, final concentration = 0.01 M, so [X] = 0.01 – 0.50 = –0.49 M. The time change is 1.5 minutes (or 90 s). So the rate can be either

   rate

   =

   –0.49 M 1.5 min

   =

   –0.33 M/min

   rate

   =

   –0.49 M 90 s

   =

   –5.4 x 10–3 M/s

   rate = –0.49 M/1.5 min = –0.33 M/min

   rate = –0.49 M/90 s = 5.4 x 10^–3 M/s

2. It is implied that there was no product to start, so the initial concentration = 0 M. Thus [X] = 0.056 M. The change in time is 75 s. So rate = 7.5 x 10^–4 M/s.

3. [X]₀ represents initial concentration. Therefore average rate =
   (0.34 – 1.00)/(50 – 0) = 0.013 M/s.

>> Instantaneous Rates

In a graph of concentration versus time, rate is the slope. However, this graph is not usually linear. Consequently, to determine rate from a graph of concentration versus time, a tangent line at the relevant time is used.

>> Example 2

What is the instantaneous rate at 1 minute and 3 minutes, based on the figure below?

Solution:

To determine the instantaneous rate, draw a line tangent to the curve at the relevant time value.

Determine the slope of the line from n = (y₁ – y₂)/(x₁ – x₂). The slope is the rate of reaction. For 1 minute your value should be about 0.1 M/min. Your tangent line and points chosen to determine the slope may be slightly different. The process is not very exact. For 3 minutes your value should be about 0.03 M/min.

>> Initial Rates

Initial rates are the rate at time zero, or when the product concentrations are zero. These are calculated in the same way as instantaneous and average rates.

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Because the production of product and consumption of reactant are related by stoichiometry, the rate relationships are too. Assuming the reaction aA + bB cC + dD, where the uppercase letters represent chemical substances and the lowercase letters their stoichiometric coefficients, the relationship between the rates of reactants and products is

–1 a

[A] t

=

–1 b

[B] t

=

+1 c

[C] t

=

+1 d

[D] t

The greatest difficulty in using this equation is remembering that the entire term ([A]/t) refers to the rate. It is tempting to want to split this into concentration and time terms, but this is not appropriate. It might be easier to consider the equation as

–1 a

rateA

=

–1 b

rateB

=

+1 c

rateC

=

+1 d

rateD

The signs (+ and –) indicate direction. Sometimes direction is useful in discussing rate, sometimes the absolute value is sufficient. Consider the context and use the direction or absolute value as appropriate.

>> Example 3

For the reaction, 2 H₂ + O₂ 2 H₂O, if the rate of disappearance of hydrogen gas is –0.012 M/s, what is the rate for the disappearance of oxygen and the appearance of water?

Solution:

Using the equation

(–1/2)(rate_H2) = (–1/1)(rate_O2) = (1/2)(rate_H2O)

so

(–1/2)(–0.012 M/s) = (–rate_O2)

–0.0060 M/s = rate of O₂

and

(–1/2)(–0.012 M/s) = (1/2)(rate_H2O)

0.012 M/s = rate of water

>> Example 4

For the reaction P₄ + 6 Cl₂ 4 PCl₃, if the rate of disappearance of chlorine gas is 0.237 M/s, what is the rate of the disappearance of phosphorus and the appearance of phosphorus trichloride?

Solution:

Use the relationship –(1/a)(rate_A) = –(1/b)(rate_B) = +(1/c)(rate_C) = +(1/d)(rate_D) and make it specific for

–(rate P₄) = –(1/6)(rate Cl₂) = (1/4)(rate PCl₃)

Since the ate given is for a reactant, it is (by definition) negative. Using the value given in the problem for the rate of Cl₂ disappearance,

–(rate P₄) = –(1/6)(–0.237 M/s)

rate P₄ = –0.0395 M/s

and

(1/4)(rate PCl₃) = –(1/6)(–0.237 M/s)

rate PCl₃ = (4/6)(0.237 M/s)

rate PCl₃ = 0.158 M/s

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