Exam (elaborations)

PORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAM

0 purchase

Course
CHEM 104 (CHEM104)

Institution
Portage Learning

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...

[Show more]

Preview 4 out of 32 pages

View example

Uploaded on January 20, 2025
Number of pages 32
Written in 2024/2025
Type Exam (elaborations)
Contains Questions & answers

portage learning chem 104 module 1 module 6 exam

DREAMS Member since 9 months 116 documents sold

$13.49

Add to cart

Save

100% satisfaction guarantee
Immediately available after payment
Both online and in PDF
No strings attached

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

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller DREAMS. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $13.49. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

71241 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 15 years now

Start selling

Seller

Summary ·

Exam (elaborations) ·

Exam (elaborations)

PORTAGE LEARNING CHEM 104 MODULE 1 –MODULE 6 EXAM

Document information

Subjects

Written for