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RE: #Chemistry Challenge 5
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?
Nah.... sorry for getting so many typo
[CO2] = 1596.78 mol / 3.92 x 10 4 dm3
= 4.0734 x 10-2 mol dm-3
But the rest should be the same
:)
This result is the (lower) concentration of CO2 when it was emitted as a gas already ... but gasoline is a liquid ...
Ya, so why I am wondering what unit you are expecting
Speed of reaction: defined as how quickly or slowly a reaction takes place
(So my very initial attempt was in liter per hour)
As you wrote yourself, reaction speed is measured in v = ∆c/∆t (and if it is an educt: v = - ∆c/∆t, because ∆c is negative, so with the additional minus v itself stays positive).
But as you said, the gasoline is a liquid, and the concentration of a liquid is constant, so ∆c would be zero, so you expect v to be zero in my answer?
Let's calculate the concentration c of isooctane at t = 0 (using the volume at t = 0), and then consider how fast n(isooctane) decreases. Even if the liquid is transformed into gas you can calculate as if the volume would stay constant (the transformation from liquid into gas doesn't influence the decrease rate of isooctane).