##### please answer these simple chem questions w/ short explanation :)

*label*Chemistry

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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-] **

(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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