Exam (elaborations)

CHEM 104 Chemistry Module 1, 2, 3, 4, 5, 6 Exam (Latest-2023)/ CHEM104 Chemistry Module 1, 2, 3, 4, 5, 6 Exam/ CHEM104 Module 1 to 6 Exam: Portage Learning |100% Correct Q & A|

19 views 2 purchases

Course
CHEM 104: General Chemistry II w/Lab

Institution
Portage Learning

CHEM 104 Module 1 Exam, CHEM 104 Module 2 Exam, CHEM 104 Module 3 Exam, CHEM 104 Module 4 Exam, CHEM 104 Module 5 Exam, CHEM 104 Module 6 Exam (Latest-2023): Portage Learning |100% Correct Q & A| CHEM104 Module 1 Exam, CHEM104 Module 2 Exam, CHEM104 Module 3 Exam, CHEM 104 Module 4 Exam, CHEM 10...

[Show more]

Preview 4 out of 31 pages

View example

Uploaded on February 22, 2023
Number of pages 31
Written in 2022/2023
Type Exam (elaborations)
Contains Questions & answers

chem 104 module 1 exam
chem 104 module 2 exam
chem 104 module 3 exam
chem 104 module 4 exam
chem 104 module 5 exam
chem 104 module 6 exam latest 2023 portage learning |100 correct q amp a|

Institution
Portage Learning
Course
CHEM 104: General Chemistry II w/Lab

DRKELVIN

Member since 3 year 610 documents sold

105

CA$36.81

Added

Add to cart

Add to wishlist

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

Module 1:

Question 1
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data table
is obtained:
2 N2O5 (g) → 4 NO2 (g) + O2 (g)

Data Table #2

Time (sec) [N2O5] [O2]

0 0.300 M 0

300 0.272 M 0.014 M

600 0.224 M 0.038 M

900 0.204 M 0.048 M

1200 0.186 M 0.057 M

1800 0.156 M 0.072 M

2400 0.134 M 0.083 M

3000 0.120 M 0.090 M

1. Using the [O2] data from the table, show the calculation of the instantaneous rate early in the
reaction (0 secs to 300 sec).
2. Using the [O2] data from the table, show the calculation of the instantaneous rate late in the reaction
(2400 secs to 3000 secs).
3. Explain the relative values of the early instantaneous rate and the late instantaneous rate.

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

2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10-5 mol/Ls
3. The late instantaneous rate is smaller than the early instantaneous rate.

,Question 2
The following rate data was obtained for the hypothetical reaction: A + B → X + Y
Experiment # [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].
2. Determine the reaction order with respect to [B].

3. Write the rate law in the form rate = k [A]n [B]m (filling in the correct exponents).
4. Show the calculation of the rate constant, k.

Your Answer:
rate = k [A]x [B]y

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

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

0.25 = 0.5x
x=2

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

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

0.125 = 0.5y
y=3

rate = k [A]2 [B]3

2.0 = k [0.50]2 [0.50]3
k = 64

Question 3
ln [A] - ln [A]0 = - k t 0.693 = k t1/2
An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day
sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the
paper.

,Your Answer:

0.693 = k t1/2

0.693 = k (5720)

k = 1.21 x 10-4

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

ln 19.8 - ln 100 = - 1.21 x 10-4 t
t = 13, 384 years

Question 4
Using the potential energy diagram below, state whether the reaction described by the diagram is
endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your answer.

Your Answer:
The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it
has relatively large Eact.

Question 5
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO
and 2.40 mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of
H2O and corresponding amounts of CO, H2, and CH4.

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

Your Answer:

0.309 mole of H2O formed = 0.309 mole of CH4 formed

, 0.309 mole of H2O formed = 0.800 - 0.309 = 0.491 mole CO

0.309 mole of H2O formed = 3 x 0.309 mole H2 reacted = 2.40 - (3 x 0.309) = 1.473 mole H2

[CO] = 0.491 mole / 8.00 L = 6.1375 x 10-2 M

[H2] = 1.473 mole / 8.00 L = 18.4125 x 10-2 M

[CH4] = 0.309 mole / 8.00 L = 3.8625 x 10-2 M

[H2O] = 0.309 mole / 8.00 L = 3.8625 x 10-2 M

Kc = [3.8625 x 10-2] [3.8625 x 10-2] / [6.1375 x 10-2] [18.4125 x 10-2]3

Kc = 3.89

Question 6

Explain the terms substrate and active site in regard to an enzyme.

Your Answer:

Enzymes are large protein molecules that act as catalysts and increases the rate of a reaction. The
catalysts act only on one type of substance to cause one type of reaction and this is called a substrate.
Active sites are enzymes that are spherical in shape together ith a group of atoms on a surface of
protein. These active sites bind with the substrate causing it to undergo a reaction. The substrate and
active site forms a complex that creates a new pathway with lower activation energy and thus speed up
a reaction.

Question 7
The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of
predominately reactants, predominately products or significant amounts of both products and reactants.
Be sure to explain your answer.

2 O3 (g) 3 O2 (g) Kc = 2.54 x 1012

Your Answer:

Kc = 2.54 x 1012 is very large. Thus, the equilibrium mixture is made up of predominantly products.

Question 8
The equilibrium reaction below has the following equilibrium mixture concentrations:

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 DRKELVIN. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for CA$36.81. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

79373 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling

Exam (elaborations)

CHEM 104 Chemistry Module 1, 2, 3, 4, 5, 6 Exam (Latest-2023)/ CHEM104 Chemistry Module 1, 2, 3, 4, 5, 6 Exam/ CHEM104 Module 1 to 6 Exam: Portage Learning |100% Correct Q & A|

Document information

Subjects

Written for

Seller

Reviews received

Content preview