100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
CHEM 104 LATEST EXAM MODULE 1-6 PRIORITY EXAM 2024 WINTER-SPRING QTR $14.59   Add to cart

Exam (elaborations)

CHEM 104 LATEST EXAM MODULE 1-6 PRIORITY EXAM 2024 WINTER-SPRING QTR

1 review
 0 view  0 purchase
  • Course
  • Institution

CHEM 104 LATEST EXAM MODULE 1-6 PRIORITY EXAM 2024 WINTER-SPRING QTR

Preview 4 out of 34  pages

  • February 19, 2024
  • 34
  • 2023/2024
  • Exam (elaborations)
  • Questions & answers

1  review

review-writer-avatar

By: LECTNAVAL • 3 months ago

avatar-seller
CHEM 104 LATEST EXAM MODULE 1-6
PRIORITY EXAM 2024 WINTER-SPRING
QTR


Module 1:

QUIZ 1
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following
data tableis obtained:
2 N2O5 (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

→ 4 NO2 (g) + O2 (g)




• Using the [O2] data from the table, show the calculation of the instantaneous rate
early in thereaction (0 secs to 300 sec).
• Using the [O2] data from the table, show the calculation of the instantaneous rate late in
the reaction(2400 secs to 3000 secs).

,• Explain the relative values of the early instantaneous rate and the late instantaneous rate.


SELECTED RESPONSE:
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.

QUIZ 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

• Determine the reaction order with respect to [A].
• Determine the reaction order with respect to [B].
• Write the rate law in the form rate = k [A]n [B]m (filling in the correct exponents).
• Show the calculation of the rate constant, k.


SELECTED RESPONSE:
rate = k [A]x [B]y

rate 1 / rate 2 = k [0.50]x [0.50]y / k
[1.00]x [0.50]y2..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]y8..0 = [0.50]y /

[1.00]y

0.125 = 0.5yy = 3

rate = k [A]2 [B]3

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


QUIZ 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 thepaper.


SELECTED RESPONSE:

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


QUIZ 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 SELECTEDRESPONSE.




SELECTED RESPONSE:

, The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous
because ithas relatively large Eact.

QUIZ 5
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800
mole of COand 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)

SELECTED RESPONSE:

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

QUIZ 6

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


SELECTED RESPONSE:

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

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

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

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

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 Hosmerit. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $14.59. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

79223 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

Recently viewed by you


$14.59
  • (1)
  Add to cart