For any general reaction 2A + B <--> 3C + D
The equilibrium expression will be given by...
k = [C]^3 [D] / [A]^2 [B]
*Liquids and Solids are omitted from the equilibrium expression*
For the rxn H2CO3 <--> H+ + HCO3-
the equilibrium expression is
ka = [H+] [HCO3-] / [H2CO3]
For the rxn Ag2CrO4 <--> 2Ag+ + CrO4 2-
the equilibrium expression is
ksp= [Ag+]^2 [CrO4 2-]
By Le Chatelier's Principle we know that a rxn will shift in the direction necessary to re-establish equilibrium. Essentially, if we add something the rxn will move away from that side with the excess amount--if we take something away, the rxn will go in that direction to fill that hole.
(a) adding extra CrO4 2- to rxn 2a...
We add CrO4 2-. The chromate ion is on the product side, so we expect the reaction to shift from the product side to the reactant side (from right to left)
(b) removing H+ from rxn 2b...
We remove H+. The hydrogen ion is on the product side, so we expect the reaction to shift from the reactant side to the product side to fill the hole (left to right)
are the two bottom answers opposite? isn't product to reactant left to right and reactant to product right to left? i may be misunderstood, please confirm
3a makes sense just confused about 3b
From product to reactant will be from right to left...
Take the rxn...
R <--> P
where R is the reactants and P are the products.
Le Chatelier's Principle tells us that if we add P, the rxn will shift from P to R (right to left).
Le Chatelier's Principle tells us that if we take some P away, the rxn will shift from R to P (left to right)
You can basically think of this principle by imagining that a rxn begins at equilibrium--it has a certain ratio of P and R that is stable. Then you disturb that ratio by either adding or taking stuff away. The rxn will move in such a way to get back to that ratio .
Number wise, suppose the equilibrium ratio (P/R) were to be 2/3...
that means two parts P and three parts R are stable at equilibrium
Now suppose you dump five more parts of P...
the ratio is now 7/3 which is greater than our necessary equilibrium ratio of 2/3!
What can possibly happen to get back to a 2/3 ratio?
What occurs is some of the P (the excess P) turns into R, which might occur at the ratio 4/6, because 4/6 = 2/3.
Specifically for (3b), the rxn is H2CO3 <--> H+ + HCO3-
which we'll generalize to R <--> P for the sake of discussion
Think of it this way: suppose the equilibrium ratio (P/R) were to be 4/8 = 1/2
that means four parts P and eight parts R are stable at equilibrium
Now suppose you TAKE AWAY two parts of P...
the ratio is now 2/8 which is SMALLER than our necessary equilibrium ratio of 4/8!
What can possibly happen to return to a 4/8 = 1/2 ratio?
What happens is that some of the R turns into P (which we are deficient in), which might occur at the ratio 5/10, because 5/10 = 4/8 = 1/2
So for (3b) the rxn will shift from R to P or from left to right
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