Equil. reaction in CSSA Trial (Industrial) (1 Viewer)

[howareya]

What did people put down for the calacuation of the equil. Involving the Haber process.

I had 0.73 was that right

This is how I worked it out

.20 + .60 --> 0

at equil.

{.20 (0.5)} + {.60-(3 x 0.5)} --> .10

then you get

.10 (squared) / .45 (cubed) x .15

= 0.73

 

[Huy]

N2 + 3H2 -> 2NH3

Initially 0.2, 0.6, 0
At equil -, -, 0.10

0.8 Initially
0.1 "used"
0.7 LHS

mole ratio 1:3 -> 2
0.7 = 4
therefore 1 = 0.175
2 = 0.35
3 = 0.525

At equilibrium, then:
N2, 3H2, 2NH3
0.175, 0.525, 0.10
(ie .175 + .525 + .1 = 0.8 as initially 0.2 + 0.6 = 0.8)

K = [NH3]^2 / [N2][H2]^3

= (0.10)^2 / (0.175)(0.525)^3
= 0.01 / (0.175)(0.525)^3
= 0.39

that's what I got ... don't know if it's right or not, it's probably wrong

 

[howareya]

i dont think you split it up like you:

mole ratio 1:3 -> 2
0.7 = 4
therefore 1 = 0.175
2 = 0.35
3 = 0.525

you have to minus it from each

 

[Huy]

I don't know,
I'm sure you had to fill in the table to work out the equilibrium constant.

Because they had:

N2 - 3H2 - 2NH3
0.2 - 0.6 - 0
--- - --- - 0.10

I must have screwed it up, but you're right, you're supposed to minus it from the initial values.

I'm not too sure about your "3 x 0.5" step.
Because you can't really say that 0.5 was used in equal proportions in N2 and 3H2, just because 0.10 of 2NH3 was used.

Maybe I don't know what I'm on about, but I went off my mole ratios of 1:3 -> 2.

Any other takers?

 

[howareya]

look in the excel book

 

[Huy]

Hehe okay, that's probably why, I *never* use the Excel Chemistry text, only Conquering Chemistry (Roland Smith).

I'll have a looksy.

 

[mat]

your right the ans is 0.73

 

[howareya]

YESSSS!!!!!!

 

[howareya]

what did other people get?

 

[sneaky pete]

I got 0.0(something)..
I totally messed up the industrial paper33