Destroyer Gen Chem # 64, Calculating Kp

[mestrada9447]

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2A + B -> 2C

0.5 atm of A + 0.2atm of B are placed in a flask @ 300K. At equilibrium Ptot= 0.6. Calculate Kp.


Isn't there a faster way to solve this without having to draw out an entire ICE table?


Destroyer is really kicking my butt.

 

[2PacClone23]

Is this 2009 or 2011? When's your test?


I actually don't even know how to do this lol.

 

[mestrada9447]

2009. The answer is 4.4

Kp= [C]^2/[A]^2



Here is the ICE table:

-------------------2A ----- +----- B <---> ------ 2C
Initial: ------------0.5------------ 0.2------------- --
Change:-------- ( -2X)------ -- (-X) ----------+2X
Equilbm:------- 0.5-2X------- 0.2-X ------------ 2X

Ptot= (.5-2X) + (0.2-X) + (2X)
Ptot= 0.6

SO

0.6=(.5-2X) + (0.2-X) + (2X)
X=0.1

THEN

[A]= 0.5-2(0.1)=0.3atm
= 0.2-(0.1)=0.1atm
[C]= 2(0.1)= 0.2atm

Kp= (.2)^2 / (.3)^2 (.1) = 4.4