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### Topic: equalib prob 2  (Read 3743 times)

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#### ashee805

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##### equalib prob 2
« on: March 02, 2005, 09:30:25 PM »
hello, here's another equalb. prob...

at a certain temp an equalib. mixture of

NO2 + SO2 ------>  NO + SO3

in a 2.0 liter container  (all are gases) is analyzed and found to contain the following:
0.200 mole NO2
.600 mole SO2
4.00 mole NO
1.200 mole SO3

then, .500 mole SO2 is injected into the container at the same temp. when equalib is reestablished, what are the new concentrations of NO2 SO2 NO aand SO3?

here is what i did:
NO2 + SO2 ------>  NO + SO3
I .200    .600            4.000  1.200

C -X      -X                +X       +X

E (.200-X) (.600-X)    (4.00+X)   (1.200+x)

do i divide them all by 2 in the initial stage or at equalib. and  what do i do now?

#### AWK

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##### Re:equalib prob 2
« Reply #1 on: March 03, 2005, 02:35:38 AM »
The first part allows you to calculate the equilibrium constant  Kc (you can use moles instead of concentration because sums of moles of reagents for both side are equal.