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RE: #Chemistry Challenge 5
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? :)
Well, with this little help, we shall get v = - (delta c)/(delta t) = - (-6.579 mol/l)/21888 s = 3 x 10^(-4) mol/(l*s)
(sorry for the bad formatting)
Congrats @thepe! Finally you are the one who found the solution of part 4! Congrats also to @bluejay188 and @mcw who solved parts 1 to 3 and worked hard to find the solution of part 4 (and nearly solved it as well). :)