Could anybody help me to understand this Q ?

BaSO4 <........> Ba^2+ + SO4^2- , △H >0

In which direction will the equilibrium shift if

a) H2SO4 is added ? Why ?

b) BaCl2 is added ? Why ?

c) NaCl is added ? Why ?

d) Heat is added ? Why ?

because it's endothramic it will shift to the right .



HNO2 + H2O <.........> H3O^+ + NO^2

In which direction will the equilibrium shift if

a) NaOH is added ?

b) NaNO2 is added ?

c) HCl is added ?

d) The acid solution is made more dilute ?



On the basis of Le chatelir principle , explain why Ag2CO3 dissolves when HNO3 is
added ?


Thanks advance

Le Chatelier's principle states that a system in equilibrium will remain in equilibrium unless a change is brought about. The system will then try to oppose the change.

I'll do the first one, you'll try the others.

1. a) H2SO4 will provide more SO4 2- ions, so the equilibrium will shift to the left so as to decrease the number of SO4 2- ions.

b) BaCl2 will provide more Ba 2+ ions, so the equilibrium will shift to the left so as to decrease the number of Ba 2+ ions.

c) NaCl has no effect on the reaction, since, there are none of Na+ or Cl- ions involved.

d) The reaction is endothermic. Of heat is added, the system will try to cool down. So, the forward reaction is favoured.

Ok thanks for your helping in first one , and I will try to solve the second Q

Could you check ?

a) no effect

b) shift to the right to decrease the # of NO2

c) no effect because no H+ no Cl-

d) I can't solve it


And the last Q , What about it

Hmmm...

2 a) you have to find if the NaOH reacts... here the NaOH will react with the acidic HNO2. Therefore, adding NaOH will decrease the number of moles of HNO2 on the left hand side. So, equilibrium shifts to the left.

d) How do you make an acid dilute? By adding water.

3 the equation is



Carbon dioxide is always being removed as it is insoluble in water. So, what can you conclude?

d) That's mean it will shift in the right..

3) I didn't understand

For d) yup, you understood.

3.

Carbon dioxide is insoluble in water. In the system, therefore, you'll always have carbon dioxide escaping. As a result the equilibrium shifts to the right. That's why you don't have silver carbonate back, unlike the previous reactions where the reactants could reform. So, the silver carbonate will always be reacting with the acid forming silver nitrate, which appears as if the silver carbonate was dissolving.

The Chatelier's principle here applies as you remove the carbon dioxide The system will try to oppose the change, by producing more carbon dioxide, that is by shifting to the right. Of course, no more reaction will occur when all the silver carbonate is used up.

Ok? :)