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### Topic: equalib *delete me*  (Read 3729 times)

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#### ashee805

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##### equalib *delete me*
« on: March 01, 2005, 08:39:57 AM »
calculate the equal constant for the following reaction:

2A -----> B + 3C  all gases

given that a 1.0 vessel was initially filled with 5.0 atm of pure A and the partial pressure of gas A was fpund to be 3.5 atm at equalib.

i did this:

2A ----->  B + 3C
I 5.0          0    0

C -2x         +x  +3x

E 5-2X         X     3X

A=  5-2(3.5) = -2

I SET IT UP AND SOLVED FOR Kc  but didnt get the answer which, btw, is 0.70 atm.  neeeed *delete me*  thanks.

#### Donaldson Tan

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##### Re:equalib *delete me*
« Reply #1 on: March 01, 2005, 10:07:57 AM »
in cases of gas, it's called Kp , not Kc, because you use partial pressure instead of concentration to calculate the equilibrium constant.

your ICE table is right, but not your calculation. let x be the partial pressure of B.

partial pressure of A at eqbm = 5 - 2x = 3.5atm

2x = 5 - 3.5 = 1.5atm
x = 0.75atm

partial pressure of B = x = 0.75atm

partial pressure of C = 3x = 2.25atm

Kp = (C3*B)/A2 =  (2.253*0.75)/3.52 = 0.6973 ~ 0.70atm
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006