4.5 Equilibria
Posted by: intechemistry on: September 23, 2010
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6 Responses to "4.5 Equilibria"
August 7, 2011 at 12:18 am
Salam Sakinah.
Yes you are right.
Quick note: Sometimes with pressure changes, the eqm cannot shift if total no. of gaseous moles on LHS = total no. of gaseous moles on RHS
May 19, 2012 at 9:40 pm
Sir, me again. 
Unit 4 question:
CH4 (g) + 6H2O (s) [CH4(H2O)6] (s) methane hydrate
Write an expression for Kp for this equilibrium.
THe answer given is 1/p(CH4). But all the while we learn is Kp = p(CH4) in this case, isnt it? Or I’m wrong all the while. You know. Sleep deprived. THanks for helping meeee! 
May 20, 2012 at 11:59 am
Hello “me again” / sdp
All equilibrium constants are equation specific.
You must specify the equation you are talking about in ordr to make th corret K equation.
So if CH4 (g) + 6H2O (s) ⇌ [CH4(H2O)6] (s)
Then indeed, Kp = 1/p(CH4)
however if you were talking about this equation:
[CH4(H2O)6] (s) ⇌ CH4 (g) + 6H2O (s)
then Kp = p(CH4)
Check the equation you were given previously.
May 20, 2012 at 2:36 pm
The equation given is the first one. so the marking scheme is right yea? Ok. noted. THanksssss
May 20, 2012 at 5:51 pm
Yeah, it’s correct. Mark schemes are more correct than anything else. If in doubt, go with the mark scheme.
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August 6, 2011 at 9:59 pm
salam sir,
i am sakinah from 11m3.
i want to make some clarification about my understanding here,
it is true that when we consider the change of temperature, equilibrium position shifted whether to the left or to the right will depend on whether the rxn is exo or endo and whether temp increase or decrease.
but, if we change the conc of reactant or pressure, the equilibrium posn will change regardless of whether the rxn is exo or endo as long as it oppose the change according to LeC principle.
am i correct? thank you