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PORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAM

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PORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAMPORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAMPORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAMPORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAMPORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAMPORTAGE LEARNING CHEM 104 MODULE 1 –MODULE...

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  • January 20, 2025
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DREAMS
PORTAGE LEARNING CHEM 104
MODULE 1 –MODULE 6 EXAM

, Portage Learning CHEM 103 Module 1 Exam: tj tj tj tj tj tj




Question 1 tj




In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data table is
tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj




tj obtained:
2 N2O5 (g) → 4 NO
tj
t j tj + t j t j O2 (g)
2 (g) tj
tj t j t j tj




Data Table #2
tj tj




Time (sec) tj
[N2O5] [O2]

0 0.300 M tj 0

300 0.272 M tj 0.014 M tj




600 0.224 M tj 0.038 M tj




900 0.204 M tj 0.048 M tj




1200 0.186 M tj 0.057 M tj




1800 0.156 M tj 0.072 M tj




2400 0.134 M tj 0.083 M tj




3000 0.120 M tj 0.090 M tj




1. Using the [O2] data from the table, show the calculation of the instantaneous rate early in the
tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj




tj reaction (0 secs to 300 sec).
tj tj tj tj tj




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




tj (2400 secs to 3000 secs).
tj tj tj tj




3. Explain the relative values of the early instantaneous rate and the late instantaneous rate.
tj tj tj tj tj tj tj tj tj tj tj tj tj




YourAnswer: tj




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




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




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

,Question 2 tj



The following rate data was obtained for the hypothetical reaction: A
tj tj tj tj tj tj tj tj tj t j t j t j + t j t j B t j t j → t j t j X t j t j + t j Y
Experiment # tj [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]. tj tj tj tj tj tj tj




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




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




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




Your Answer: tj




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




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




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




0.25 = 0.5x tj tj




x =2 tj tj




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




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




0.125 = 0.5y tj tj




y=3 tj tj




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




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




k = 64tj tj




Question 3 tj



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




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




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




tj paper.

, Your Answer: tj




0.693 = k t1/2 tj tj tj




0.693 = k (5720) tj tj tj




k = 1.21 x 10-4
tj tj tj tj




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




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




t = 13, 384 years
tj tj tj tj




Question 4 tj



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



or exothermic and spontaneous or nonspontaneous, being sure to explain your answer.
tj tj tj tj tj tj tj tj tj tj tj tj




Your Answer: tj



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




relatively large Eact.
tj tj tj




Question 5 tj



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




tj 2.40 mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of
tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj tj




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




CO (g) tj
t j t j
+ t j 3 H2 (g) tj
tj t j t j
CH4 (g) tj t j
+ t j t j H2O (g) tj




Your Answer: tj




0.309 mole of H2O formed = 0.309 mole of CH4 formed
tj tj tj tj tj tj tj tj tj tj

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