kashyranz

2+ Year Member
May 27, 2015
85
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Status
Pre-Dental
I'm having trouble with when to use coefficients in equilibrium expressions.
For example in the Haber Process:
N2 (g) + 3H2 (g) --> 2NH3 (g)
the equilibrium expression would be
(NH3) ^2 over
(N2) (H2)^3
this does not put the coefficient in front and raise it all to the exponent like molar solubility problems. Can someone clarify why this is and which types of problems use the coefficient and exponent, and which use just the exponent?
 

bigbutrealdreams

2+ Year Member
Aug 24, 2015
223
99
i simply remembered that, in most problems (though not always),

for Keq expression you do not include the coefficient,
for molar solubility (Ksp) you do include them.
 

Graffix

2+ Year Member
May 12, 2015
140
87
Status
Pre-Dental
If I am not mistaken, we usually wouldn't use coefficients in equilibrium expressions because we can't have a "concentration of 2 something". We wouldn't say "I have .1M of 2NH3" . We use coefficients with molar solubility because we usually don't know the change in concentrations and the final concentrations. If you write out an ice table and the change is "+2x" or "+3x" then thats including the coefficients in the keq expression.
 

enamel88

5+ Year Member
7+ Year Member
Mar 24, 2012
519
209
CA
Status
Dental Student
usually in Ksp for instance, you don't do the coefficient in front if you're going from solutes and trying to see if it'll form a precipitate. In the haber process, i think you're going from solutes (gas components) to the Ammonia gas product. Check the problems where it asks you "will it form a precipitate or not".