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Chemistry 7 Online
OpenStudy (anonymous):

When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay: HCOOH(g) →CO2(g) + H2 (g) The rate of reaction is monitored by measuring the total pressure in the reaction container. Time (s) . . . P (torr) 0 . . . . . . . . . 220 50 . . . . . . . . 324 100 . . . . . . . 379 150 . . . . . . . 408 200 . . . . . . . 423 250 . . . . . . . 431 300 . . . . . . . 435 At the start of the reaction (time = 0), only formic acid is present. What is the formic acid pressure (in torr) when the total pressure is 291? half-life in sec? What is the rate constant (in s-1)?

OpenStudy (anonymous):

pressure of formic acid=149torr,rateconstant=0.012s-1

OpenStudy (abb0t):

You can tell by the units that the rate order is 1 (s\(^{-1}\))

OpenStudy (anonymous):

but how did you get the answer? can you please write the steps thank you.

OpenStudy (anonymous):

did u ask me?

OpenStudy (anonymous):

do you know integral form of first order rate equation

OpenStudy (anonymous):

[A] = [A]oe-kt ?

OpenStudy (anonymous):

yes using that you can find k that is rate constant

OpenStudy (anonymous):

take hcooh is A then[A]o=220torr

OpenStudy (anonymous):

at50sec let HCOOH is xtorr used up then final HCOOH=220-x finalCO2=x,finalH2O=x but given total pressure is 324 torr at 50sec so220-x+x+x=324 x=104 [A]=220-x= 116 use them

OpenStudy (anonymous):

if the total pressure is 291torr then 220-x+x+x=291 x=71 pressure of HCOOH=[A]=220-x=149

OpenStudy (anonymous):

in such problems always assume x amount is used up and continue with given data

OpenStudy (anonymous):

since one mole ofHCOOH gives one mole each of CO2 andH2O xamount also gives in same ratio

OpenStudy (anonymous):

now is that clear to you shushu55

OpenStudy (anonymous):

plz reply

OpenStudy (anonymous):

oh i see so since Ao=220 I form an equation and set it equal to the give total pressure, in this case 291. Then I find x by solving the equation which is 71 & then again 220-71=149. Yep I think I got it thank you.

OpenStudy (anonymous):

you're smart got it very fast

OpenStudy (anonymous):

it took me 1month to find that technique

OpenStudy (anonymous):

lol thanks:) you just explained it well. Trust me I'm lost in chemistry.

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