kinetics answers

Hera are the answers to the Kinetics problems, the superscripts and subscripts don't translate well, neither did the diagram. If you need clarification then I can email you the answers.

3. Tripling the [OH] causes the rate to triple so the order with respect to OH is 1
4. Doubling the [Cl2] causes the rate to double, doubling the [Cl2] and doubling the [NO] causes the rate to increase 8 fold so the rate law would be:
Rate = k[NO]2[Cl2]
5. Insert the given data for say expt. 1:
0.269M min-1 = k [0.5M] [1.8M]2
solve for k gives 0.269 M min-1/ 0.5M x 1.8M2
k = 0.166M-2 min-1, however we are asked for units of L2 mol-2 min-1 M = mol l-1
so, k = 0.166 L2 mol-2 min-1
Intergrated rate laws
1. Half life does not depend on the initial concentration for a first order reaction
1/8 of original value = 3 half lives so (500s x 3) = 1500s

2. Initial concentration [A]o = 1.00M, at t= 480s the concentration of B was 0.385M, so the concentration of A at time t [A]t is 1-0.385 = 0.615M
For a first order reaction:
ln[A]t = -kt + ln[A]o ln(1) =0

ln[A]t = -kt

ln 0.615 = -k x 480s

so k = 0.001s-1

3. For a first order reaction t 1/2 = 0.693/k
If k = 1.01 x 10-3 s-1 then t1/2 = 0.693/1.01 x 10-3
t 1/2 = 686s
4. After 10 half lifes the remaining sample = 1/210 = 1/1024 = 0.000977


Reaction mechanisms and rate laws, reaction pathways
1.

1. HBr(g) + O2(g) → HOOBr(g)
2. HOOBr(g) + HBr(g) → 2HOBr(g)
3. 2HOBr(g) + 2HBr(g) → 2H2O(g) + 2Br2(g)
a. Overall: 4 HBr(g) + O2(g) → 2H2O(g) + 2Br2(g)
b. Rate = k[HBr][O2] the rate law = the slowest, rate determining step so the rate law is the rate law for step 1. Step 1 is therefore the slow (rate determining) step
c. Intermediates are produced in 1 step and consumed in another so: HOOBr(g) and HOBr(g)
d. No, the intermediate produced in the step is consumed very rapidly in the next step

2.
a. Intermediates are produced in 1 step and consumed in another so: NO3
b. rate law is the rate law for the slow step so: rate = k[NO2]2 {uses the stoichiometric coefficients of the elementary steps (not the overall reaction)}
c.
Energy
Reaction Pathway





Something like this overall exothermic, first step big Ea endothermic, 2nd step small Ea exothermic
3. The catalyst decreases the Activation Energy. The equilibrium is not affected, it is just reached quicker.
4. Done in class:

Overall reaction:

Cl2(g) + CHCl3(g) → CCl4(g) + HCl(g)
The reactive intermediates are: Cl(g) and CCl3(g)

Rate Law is equal to the rate law for the slowest step so rate = k[Cl][CHCl3]

But Cl is a reactive intermediate so we must solve for it with something in the overall equation:

Step 1 is at equilibrium, so rate = k1[Cl2] = rate =k-1[Cl]2

So [Cl]2 = k1/k-1[Cl2]

So k1/k-1[Cl2]1/2

So overall rate law is rate = k1k2/k-1[CHCl3][Cl2]1/2 = k [CHCl3][Cl2]1/2

quiz answers

Here are some answers to the quiz, all the ones I did by hand I will need to give the answers out in class...

1. Assign names to the following hydrocarbons:






2. Draw the structure of: will do on board
i. Cis 5-methyl-3-heptene
ii. 2- Pentanol
iii. 3 Methyl cyclohexene
iv. Trans 2-methyl-2-hexene
v. 2-methyl 1,4-hexadiene

3. i. Arrange the following alcohols in order of INCREASING boiling point

CH3OH CH3CH2CH2OH CH3CH2CH2CH2OH HOCH2CH2CH2CH2OH
ii. Explain your reasoning
All alcohols exhibit hydrogen bonding. Those with more than 1 OH group will experience greater hydrogen bonding.
Bigger molecules are more easily polarized and hence experience greater London dispersion forces

4. Explain why we ‘recrystallized’ our synthesised Xylene Sulfonic Acid crystals
5.
To increase the purity. Crystals a re dissolved in hot solution. The solution is then allowed to cool. As the solution cools the solubility of compounds in solution drops. This results in the desired compound dropping (recrystallizing) from solution. The slower the rate of cooling, the bigger the crystals formed.

6. Which of the following alcohols could be oxidized to an aldehyde
The primary alcohol – the one with 2 C-H bonds on the C bonded to OH

7. He following ester could be made from which carboxylic acid and alcohol (draw the structures and name)



8. Balance the following equation:

3 CH3CH2OH + 2 Cr2O72- + 16 H+ → 3 CH3COOH + 4 Cr 3+ + 11 H2O

In this reaction the dichromate is oxidizing ___ethanol_______ to _______ethanoic acid__________

9. Complete the following equations with reactants or products and/or conditions






10. Draw 3 structural isomers of C3H8O try naming the 3 that you draw.
Will do on board

11. In the Organic compound CH3X, the weight % of X is 32, what is X? Lithium

12. The Lewis structure of the following organic molecule, what are the approximate bond angles at the points mark a,b and c






13. What is the hybridization of the C atoms labelled a and b in the molecule

14. Explain briefly why Phenol is more acidic than ethanol

The phenoxide ion C6H5-O- is stable as the non bonding electrons are delocalized into the ring.

15. Complete the following reaction
2Na(s) + 2C2H5OH(l) → 2C2H5ONa + H2 sodium ethoxide is a strong base (conjugate base of the ethanol(weak acid)
16. Explain why benzene does not readily undergo addition reactions
It would destabilise the planar hexagon ring structure, which is stable due to delocalisation of electron density over the molecule

17. Why will hexane not dissolve in water?
It is non-polar, water is polar

18. Carboxylic acids are weak acids, however they will readily lose a proton to form the carboxylate anion (the conjugate base). Explain why the carboxylate anion formed is a stable ion?
It can exist as 2 resonance forms. The charge can be spread over a wide area

19. Why should a reflux condenser never be plugged at the top?
The pressure build up will cause the apparatus to explode

20. Base catalysed hydrolysis of an ester is also known as ___________?
Saponification

21. Why would the Cis isomer of 2-butene have a slightly higher boiling point than trans 2-butene?
The slight dipole moments of the C-H bonds cancel out in the trans isomer but not in the cis form leading to a slight increase in Intermolecular forces.
However trans isomers pack together more efficiently in the solid form and hence tend to have slightly higher melting points

some more sample problems.

Well done for everyone who made it through the Benzoic Acid Lab on friday, remember lab books are due in by the end of tuesdays class.

Hopefully everyone has the sample set of problems, I'll post the answers next weekend. Here are a few more suggested problems to check out from the book

22.1, 22.2, 22.3, 22.4, 22.17, 22.25, 22.38, 23.13, 23.28

Also the links to the sites mentioned in the class this week are:

Wiki's naming site

http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry

A great resource for Organic Chem.. I suggest the alcohols link

http://www.chemguide.co.uk/orgpropsmenu.html#top

http://www.chemguide.co.uk/organicprops/alcohols/background.html#top

If you are planning on taking more Organic Chem in the future this is a good resource for later

http://www.chemguide.co.uk/mechmenu.html#top

Esters

http://en.wikipedia.org/wiki/Ester

This week we'll wrap up amines and amides and do Chirality on Monday

Tuesday and Thursday we'll look at Polymers.

I'll post a few more questions on these topics during the weeks.


Also the mystery of the $30 has been solved, however as promised we'll have some donuts anyway this week.

Lab

Remember if you haven't already handed in your Lab book, you have until tuesday's class to do so. If you lost $30 in the lab on friday, don't worry you can get it back in mondays class. Have a good weekend

Here are some good MSDS sites for fridays lab:
http://msds.chem.ox.ac.uk/SU/sulfuric_acid_concentrated.html
http://msds.chem.ox.ac.uk/HY/hydrochloric_acid.html
http://msds.chem.ox.ac.uk/XY/p-xylene.html

I'll write up a template for the lab notebooks at the start of the lab

Week 1

Thanks for checking out the Blog, remember that the blog can be used to post questions as well, it’s often the case that the question you have will be something that almost everyone needs clarification on.
This week we started studying Organic Chemistry- the good stuff.
We started with hydrocarbons, those compounds made up of carbon and hydrogen only.
Hopefully everyone is OK with naming relatively simple alkanes, next week we’ll name alkenes and alkynes as well, and include geometric isomerism and the cis or trans prefix, we’ll also look at some simple cyclic alkenes, in summary:
Remember, look to see what the family is (alkane, alkene, alkyne). Look for the longest chain. (not always the straight part). Find the first substitiuent (for alkanes) and start numbering the chain from that end. When 2 or more substituents are present list them in alphabetical order, if 2 or more of the same substituent are present remember the di, tri, tetra prefix. The naming principle is the same for alkenes and alkynes, except when numbering the longest chain find the end closest to the double or triple bond and start that end as 1. If there is more than one double bond, then add the di, tri, tetra to the ENE.....i.e. 1,3 hexadiene for example. The rules for cis and trans isomers are, if the main chain is all on the same side its CIS, if its on opposite sides of the double bond its TRANS. (remember to put the cis- and trans- prefix in the name). For cycloalkenes, number through the double bond in the direction to give the substituent the lowest number, this can be clockwise or counter clockwise.

Anyway, more of this next week.

The Suggested Questions from the book were: 21.41, 21.42, 21.43, 21.44, 21.45 also 22.17
Here are the links to the acetylene ‘you tubes’ that are in the notes
http://www.youtube.com/watch?v=HUVNf-y349E&feature=related
http://www.youtube.com/watch?v=45y1lSnlrH8
With respect to the Lab Safety and Write up class on Friday. Try and get hold of a lab book, (hard bound if possible), some safety specs and a lab coat (or suitably protective clothing).
Next week will be a lab book only report and I’ll give you a template style to follow before class.
The Pre-lab, must be handed in to me prior to Fridays lab. You should have read the lab before coming and looked up the relevant MSDS hazards for the chemicals you will be using. On Friday we are doing the sulphuric acid, p-xylene experiment.
Lab reports need to be handed in at or before Tuesday’s 12.30 class! Unless you have extenuating circumstances, or are otherwise instructed.

Have a great weekend and see you all next week.