#Chemistry Challenge 5
Once more I ( @jaki01 ) let run my #chemistrychallenge under the SteemSTEM flag to support this still growing science community which in my eyes adds a lot of value to our Steemit community!
What could be more fun than activating your brain and at the same time having the opportunity to win up to 30 STEEM POWER. :-)
Rules - what are the requirements to win the prize?
The winner is the first person to post the correct solution (correct number and correct unit; solution process required). The prize for the correct answer is 30 STEEM POWER. We will send it to the winner immediately after the article is paid out (7 days after posting).
If nobody is able to determine the solution before Sunday, July 9th, 2017 at 10 pm (German time!) the prize (30 STEEM POWER) will be added to the next competition (bringing that one up to 45 STEEM POWER!).
Your Chemistry Challenge to Solve consists of four parts this time:
-
What is the correct equation for the reaction of isooctane (2,2,4-Trimethylpentane, C8H18) with oxygen O2 to carbon dioxide CO2 and water H2O? (Hint: it is a redox reaction.)
-
Let's assume that gasoline consists to 100 % of isooctane C8H18, and a car would consume exactly 5 liters gasoline when it goes over a distance of 100 kilometers. One liter of gasoline would cost 1.50 euro. How many kilometers could the car go if the driver tanks up his vehicle for exactly 45.60 euro (tank was empty before)?
-
How many kg and liters of CO2 would be emitted by the car while going the distance you calculated in part 2?
Given parameters:
ρ(C8H18) ≈ 0.75 g∙cm-3 ; molar masses: C = 12 g/mol ; H = 1 g/mol ; O = 16 g/mol ; universal gas constant R = 8.314 kPa∙l∙mol-1∙K-1; temperature T = 295 K (kelvin);
air pressure p ≈ 1 bar ; 1 bar = 105 Pa
Hint: use the equation of part 1! -
Let's assume, the average speed of the car while going the distance which you calculated in part 2 was 100 kilometers per hour (100 km/h).
What is the average reaction speed of dissipating isooctane v(C8H18) during this timescale?
implies 30.4 L could go 608 km
implies mole of isooctane = 199.597 mole
implies mole of CO2 = 1596.78 mole
implies mass of CO2 emitted = 70.26 kg
and since PV = nRT; Rearrange and get
volume of CO2 emitted = 3.92 x 104 L
implies 5 L/hr
1. Your equation is correctly balanced, but I would have preferred if you use the smallest possible whole numbers as factors in front of the element symbols. I count your answer as correct though.
2. Correct answer!
3. Why didn't you use the given (rounded) molar masses? In that case we would get 'nicer' numbers as results and I wouldn't need to verify every single of your calculations.
Your mass of isooctane is correct. Also n(C8H18) and n(CO2), but the mass of CO2 (even with not rounded molar masses) is a little bit too small. What did you use as M(CO2)? The volume (calculated with your numbers) is correct again (I guess you rounded R different than given).
4. The speed of chemical reactions isn't measured by volume / time!
I am just lazy so I use chemdraw to generate the molecular weigh LOL
More explanation on
Q3)
mole of CO2 = 1596.78 mole
implies mass of CO2 = 1596.78 * 44 = 70258 g = 70.26 kg
I use the given R btw
Q4
I thought the v(C8H18) where the v means volume LOL, but now i guess probably you mean speed
I would take rate of reaction as concentration over time usually, but I don't have concentration here, so I guess you are looking for change of mole per unit time?
If this is the case, here would be my answer:
5 L/ hr implies 32.89 mol/hr
Yes, right, but first you had another volume and also another mass of CO2. Better is to add a new comment if you change any parts of your solutions.
Concerning part 4: if any information seems to be missing you can use every given detail from parts 1 to 3 to find it out.
So far your answer is wrong.
How about:
v(C8H18) = - 32.89 mol/hr
NOT correct! :-)
But apart from that so far you did well ... :)
Edit: by the way, the reaction speed v is not negative (even if ∆c is negative), but that's not the point: your number is wrong, too.
humm... if you are expecting the answer in concentration:
[CO2] = 4.734 x 10-2 mol dm-3
implies rate of CO2 production = 6.7 x 10-3 mol dm-3 hr-1
implies speed of dissipating isooctane = 8.37 x 10-4 mol dm-3 hr-1
Thanks for letting us know about the concentration of carbon dioxide.
How exactly do you get
c(CO2) = 4.734 x 10-2 mol/dm3?
m(CO2) is correct now, but please make a remark if you edit your text! (Same with your volume which was wrong at first.)
We need to know when you made your changes.
Sure Sure! Definitely !
Anyway, its seems not too many chemistry people around steemit, sadly, so it's really nice to meet you here :D
Btw, feel free to have a look my blog on some interesting chemistry sharing :D
:-)
I already follow you ...
Great! hope you enjoy it :D
Later I will check it more thoroughly!
45,60 Euro Get 3.40 L of Isooctane (45.6 / 7.50 * 100 = 608 KM)
@mcw correct i come late
1.) 2 C8H18 + 25 O2 ===> 16 CO2 + 18 H2O
2.) 608 km (Dimensional Analysis)
3.) 70.3 kg CO2
3.9 x 10^4 L CO2 (Dimensional Analysis / Ideal Gas Law)
4.) .306 [C8H18]/hr (dc/dt)
Very nice, even if @mcw has solved parts 1 to 3 already as well.
Concerning part 4 you are both wrong (your idea is good though ... just the number is wrong). You may show me your exact calculations ... so I can see where the difference comes from ... for example, what did you chose as ∆c and ∆t? :)
By the way, I would prefer the unit mol∙l-1∙s-1
9.11 x 10^-4 mol∙l^-1∙s^-1
Ok, I think I've figured it out. So the Rate equation for this would be Rate = ((1/2)(d[C8H18]/dt)).
You can find d[C8H18] by dimensional analysis from 70.3 kg CO2... which is about 199.7 moles of C8H18... divide that by 5 to get the concentration which is 39.9 M. Now, just divide that value by the time elapsed for the car to go 608 km, which usind DA should be 21888 s. Now, if we plug those into our original equation, the value should be 9.11 x 10^-4 mol∙l^-1∙s^-1. Let me know if that needs to be fixed anywhere - thanks! :)
Time (21888 s) is correct, but ∆c isn't.
Why would you divide the 200 mol (it is 200 mol, if you use the given molar masses) of isooctane by 5?
Ah, that's how many Liters are required for 100 km, yet we're going 608 km. SO instead, 200 should be divided by 30.4 making the answer 1.50 x 10^-4 mol∙l^-1∙s^-1
Sounds really good now - the only thing I wonder about is why you are using the factor 0.5?
Don't you need to use it to account for the molar ratios of the balanced equation?
No, if we had 6.579 mol/l isooctane at the beginning and 0 after 6.08 h, than ∆c in the timescale was - 6.579 mol/l.
So what is v now finally? :)
Good challenge.
But may I make a suggestion: there is now a Chemistry forum in chainBB and would be good to include any chemistry challenges within it - https://beta.chainbb.com/forum/chemistry - but in order to appear there each post must have #chemistry tag or you can make a request to have, for example, #chemistrychallenge as a tag for the forum. Just an idea.
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The contest tag has been removed and replaced with #chemistry. It is important to maintain accessibility through both steemit and chainbb.
thank you. i wish i can answer.
AHHhhh I'm always late to these :(
Imma be on guard for a biology challenge... that will be my time!!!
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