PORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAM Compiled

PORTAGE LEARNING CHEM 104 MODULE 1
–MODULE 6 EXAM Compile



PortageLearningCHEM103Module1Exam:q q q q q q




Question 1 q




In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data table is
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qobtained:
2 N2O5 (g) → 4 NO
2 (g) + O2 (g)
q

q q q q

q q q q




DataTable#2
q q




Time(sec) q
[N2O5] [O2]

0 0.300 M
q 0

300 0.272 M
q 0.014 M q




600 0.224 M
q 0.038 M q




900 0.204 M
q 0.048 M q




1200 0.186 M
q 0.057 M q




1800 0.156 M
q 0.072 M q




2400 0.134 M
q 0.083 M q




3000 0.120 M
q 0.090 M q




1. Using the [O2] data from the table, show the calculation of the instantaneous rate earlyin the
q q q q q q q q q q q q q q q q

,reaction (0 secs to 300 sec).
q q q q q




2. Using the [O2] data from the table, show the calculation of the instantaneous rate late in the reaction (2400
q q q q q q q q q q q q q q q q q q




qsecs to 3000 secs).
q q q




3. Explainthe relative values of the earlyinstantaneous rate and the late instantaneous rate.
q q q q q q q q q q q q q




YourAnswer:
q




1. rate = (0.014 - 0) / (300 - 0) = 4.67 x 10-5 mol/Ls
q q q q q q q q q q q q q




2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10-5 mol/Ls
q q q q q q q q q q q q q




3. The late instantaneous rate is smaller than the early instantaneous rate.
q q q q q q q q q q

,Question 2 q




The following rate data was obtained for the hypothetical reaction: A + B → X + Y
q q q q q q q q q q q q q q q q q q q q q




Experiment # q [A] [B] rate
1 0.50 0.50 2.0
2 1.00 0.50 8.0
3 1.00 1.00 64.0

1. Determine the reaction order with respect to [A]. q q q q q q q




2. Determine the reaction order with respect to [B]. q q q q q q q




3. Write the rate law in the form rate = k [A]n [B]m (filling in the correct exponents).
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q q

q q q q




4. Show the calculation of the rate constant, k. q q q q q q q




Your Answer: q




rate = k [A]x [B]y q q q
q




rate 1 / rate 2 = k [0.50]x [0.50]y / k [1.00]x [0.50]y
q q q q q q q
q q

q q
q




2..0 = [0.50]x / [1.00]x
q q q q
q

q




0.25 = 0.5x q q




x=2 q q




rate 2 / rate 3 = k [1.00]x [0.50]y / k [1.00]x [1.00]y
q q q q q q q
q q

q q
q




8..0 = [0.50]y / [1.00]y
q q q q
q

q




0.125 = 0.5y q q




y=3 q q




rate = k [A]2 [B]3 q q q q




2.0 = k [0.50]2 [0.50]3
q q q q




k = 64
q q




Question 3 q




ln [A] - ln [A]0 = - k t
q q q q
q
q q q 0.693 = k t1/2 q q q




An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day sample.
q q q q q q q q q q q q q q q q q q




The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the paper.
q q
q
q q q q q q q q q q q q q q q q q q

, Your Answer: q




0.693 = k t1/2 q q q




0.693 = k (5720) q q q




k = 1.21 x 10-4
q q q q




ln [A] - ln [A]0 = - k t
q q q q q q q q




ln 19.8 - ln 100 = - 1.21 x 10-4 t
q q q q q q q q q
q




t = 13, 384 years
q q q q




Question 4 q




Using the potential energy diagram below, state whether the reaction described by the diagram is
q q q q q q q q q q q q q q




endothermicor exothermicand spontaneous or nonspontaneous, being sure to explain your answer.
q q q q q q q q q q q q q




Your Answer: q




The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it has
q q q q q q q q q q q q q q q q q q




relatively large Eact.
q q q




Question 5 q




Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and
q q q q
q
q q q q q q q q q q q q q q




q2.40 mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of
q q q q q q q q q q q q q q q q




H2O and corresponding amounts of CO, H2, and CH4.
q q q q q q q q




CO (g) + 3 H2 (g)
q

q q
q q

q q q
CH4 (g) + H2O (g)
q q
q q q




Your Answer: q




0.309 mole of H2O formed = 0.309 mole of CH4 formed
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