1. Example Consider the decomposition of gaseous PH3:
The rate of the reaction was measured as4 PH3 (g) ---> P4 (g) + 6 H2 (g)
What is the rate in terms of (a) P4; (b) H2.
Answer: Use your knowledge of dimensional analysis to determine the other rates. Watch your signs! A negative rate means that the substance is being used up; a positive rate means that a substance is being formed. PH3 is being used up in the above example, P4 and H2 are products so their rates of formation will be positive.2. Hints See Example 15.3 in Kotz and Treichel, also Exercise 15.3 (answer in back of book). Remember:
3. Hints You have the rate equation, which relates the rate to the rate constant and the initial concentrations of A and B. You are given the rate constant and the concentrations. Solve to find the initial rate.
When half of one of the reactants has reacted, calculate how much is left of both reactants, and use these with the rate constant to find the rate at this point.
Example. For the reaction 3 A + B ---> 2 C, the rate law is rate = k[A][B], with k = 5.00 x 10-3 M-1 min-1. If the initial concentrations of A and B are 1.00 M, what is the rate when half of A has reacted?
Solution. When half of A has reacted, its concentration will be 0.50 M (1.00 M to start - 0.50 M reacted).The amount of B that has reacted at this point is (0.50 M A reacted)(1 M B/3 M A) = 0.17 M B reacted. Therefore the concentration of B when half of A has reacted = 1.00 M - 0.17 M = 0.83 M.
rate = 5.00 x 10-3 M-1 min-1 (0.50 M A) (0.83 M B) = 2.08 x 10-3 M-1 min-1
4. Hints. See p. 707 on half-life and first-order reactions.
Remember that in the integrated rate equation for first-order reactions,
5. Hints. See Section 15.1 in Kotz and Treichel, with the
example in Figure 15.3 for calculating instantaneous and average rates
of reaction. See also pp. 704-705 on graphical methods for determining
the rate constant. Figure 15.9 shows a graph of data for a first-order
reaction.
6. Hints. Refer to Figure 15.14 in your text, which shows a typical
diagram for energy changes as a reaction progresses from reactants through
an activated complex to products. The enthalpy change for the reaction
is the difference in energy between reactants and products.
7. Example. A reaction A ---> B proceeds with the product B 4 kJ/mol more stable than the reactant A. The activation energy for the uncatalyzed reaction is 262 kJ. With a suitable catalyst, the activation energy drops to 115 kJ. How many times faster is the catalyzed reaction than the uncatalyzed reaction at 227 oC?
Solution. Use the Arrhenius equation,
Set up the rate constants for the catalyzed and uncatalyzed reactions, and take their ratio to find out how many times faster the catalyzed reaction is. 227 oC = 500 oK
8. Hints. See pp. 715-718 on the Arrhenius equation, and Example
15.10.
9. Hints. The rate of a reaction is directly proportional to the rate constant. Use Equation 15.7 to determine the ratio of the rate constants at the two temperatures, then apply this ratio to the given rate to determine the reaction rate at the higher temperature.
Example. At 115 oC, a reaction produces product at the rate of 3.15 x 10-3 mol L-1 s-1. If the activation energy is 45 kJ, what will the reaction rate be at 145 oC?Solution. T1 is 115 oC, T2 is 145 oC. The reaction rates are:
Now use Eq. 15.7 to find kT2/kT1And since we know that the rate at 115 oC is 3.15 x 10-3 mol L-1 s-1, the rate at 145 oC is 2.72 times greater, or 8.57 x 10-3 mol L-1 s-1.
10. Hints.
This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem181/CHEM_1P81_Assign_6.html
Last revised: February 27 2001 by M. F. Richardson
© Brock University, 2001
