Other

[CHEM 1220]Assignment #1

55 views 0 purchase

Course
CHEM 1220 (CHEM1220)

Institution
Langara College ( )

[CHEM 1220] These are the completed assigned questions . It may look like only 4 questions, but it can easily take 8-10 hours to complete (especially since these assigned questions are for the 1st month). Question 4 involves extensive calculations and will take the longest.

[Show more]

Preview 2 out of 6 pages

View example

Uploaded on November 12, 2021
Number of pages 6
Written in 2021/2022
Type Other
Person Unknown

collision theory
iron used as a catalyst
haber bosch process
how long it takes for 9999 to decay
radioactivity after 559 hours
energetic and geometric requirement
2no2h2
n22h2o
2nog2h2g⟶n2g

RainyMonday Member since 3 year 58 documents sold

CA$15.55

Add to cart

Add to wishlist

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

Date #1: Assigned Questions Your Name

1. Fluorine-18 is a radioactive isotope that decays to form oxygen-18 with a half-life of 1.83 hr. 18F is used to study
the brain by injecting fluoro-substituted glucose into the patient’s bloodstream. The glucose accumulates in the
regions where the brain is active and needs nourishment.
a) What is the rate constant for the decomposition of fluorine-18?
b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of
the radioactivity will remain after 5.59 h?
c) How long does it take for 99.99% of the 18F to decay?

𝑙𝑛2
a) Radioactive materials are always 1st Order. Use equation t1/2= 𝑘
. Find Rate Constant first:

𝑙𝑛2 ln(2) ln(2)
t1/2= k= 1 k= k = 0.378 k = 3.78 x 10-1 1/hr
𝑘 𝑡 1.83 ℎ𝑟
2

b) This requires the Integrated Rate Law for 1st Order:

Formula: ln [A]t = ln[A]0 - k(t)

ln [F]5.59 = ln [F]0 – k(t)

ln [F]5.59 – ln [F]0 = - k (t)

[F]
ln ([F]) = - (3.78 x 10-1 1/hr) (5.59 hr)

[F]
([F]) = e-(3.78 x 10-1 1/hr)(5.59 hr) x100

= 12.1% This is how much remains after 5.59 hrs

c) If 99.99 % decays, then there is only 0.01 % left:

Formula: Use the same Integrated Rate Law again, but need time
ln [A]t = ln[A]0 - k(t)

100% - 99.99% which equals 0.01%. which is 0.0001 in a concentration amount:

ln (100%) – ln (99.99%) = - (3.78 x 10-1 1/hr)(t)

ln(0.0001) = -3.78 x 10-1 1/hr (t)
ln(0.0001)
=t
−3.78 x 10−1 1/hr

t = 24.3659 hrs It takes 24.4 hours for 99.99% to decay.

, Date #1: Assigned Questions Your Name

2. The following reaction has the reaction coordinate diagram shown below. According to collision theory, what
conditions must be met in order for the forward reaction to occur? Explain energetic and geometric
requirements.

Since this pertains to collision theory, there are 2 main requirements:

1. The molecules need to have proper orientation when colliding
2. The molecules need enough energy to overcome the reaction barrier in
order to become a product.

This reaction barrier requires the molecules to successfully surmount
the activation barrier to actually become a product. Molecules may approach the
barrier many times but may not have enough energy to get over the barrier.

3. The production of ammonia via the Haber-Bosch process is an important industrial process. The Haber-Bosch
process follows the reaction: 𝑁2(𝑔)+3𝐻2(𝑔)→2𝑁𝐻3(𝑔)
a) Explain why we can say with certainty that the reaction occurs in multiple steps
b) The reaction is usually carried out in the presence of iron. Why do you think iron is including in the reaction chamber?

a) The reactants have 4 moles reacting, and the probability of all 4 molecules colliding successfully at the same
time is very low. They would need to collide at the same time, with the proper geometry, and the right energy,
and according to collision theory, this would be extremely improbable. There would need to be at least 2
elementary steps.
b) Iron is used as a catalyst in the reaction. This speeds up the reaction between nitrogen and hydrogen gases, and
this allows for the reaction to be carried out at a lower temperature. If this occurs, more molecules collide and
more product is formed.

4. You are studying the following reaction: 2NO(g)+2H2(g)⟶N2(g)+2H2O(g)
To study this reaction in the lab, you can monitor how the concentration of NO changes over time.
The following table shows the different conditions of the experiments you performed.

Experiment # Initial [NO] M Initial [H₂] M Temperature (K)
1 0.0060 0.0010 298
2 0.0060 0.0020 298
3 0.0010 0.0060 298
4 0.0020 0.0060 298
5 0.0060 0.0010 278
6 0.0060 0.0010 288
7 0.0060 0.0010 318

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 RainyMonday. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for CA$15.55. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

56326 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

Other

[CHEM 1220]Assignment #1

Document information

Subjects

Written for

Seller

Reviews received

Content preview