I am fairly sure that a) is no change, and c) is shifts to the right...but I am not sure about b) and d). Thanks.

Question

How does the equilibrium position shift as a result of each of the following disturbances?
Sodium bicarbonate undergoes thermal decomposition according to the reaction below.
2 NaHCO3(s) <~> Na2CO3(s) + CO2(g) + H2O(g)

How does the equilibrium position shift as a result of each of the following disturbances?
(a) 0.20 atm argon gas is added.
-shifts toward the left
shifts towards the right
-no change

(b) NaHCO3(s) is added.
-shifts toward the left
-shifts towards the right
-no change

(c) Mg(ClO4)2(s) is added as a drying agent to remove H2O.
-shifts toward the left
-shifts towards the right
-no change

(d) Dry ice is added at constant T.
-shifts toward the left
-shifts towards the right
-no change

Answer 1

a and c are right.
b has no effect. You already have solid NaHCO3 present; adding a few more grams doesn't change anything. Remember solids do not show up in the keq expression. Or if you do want to include the solid, the (solid) = 1 by definition. Adding more solid and (solid) still is 1 by definition.
I suppose d, technically, may depend upon what the constant T is, but I'm inclined to think the dry ice will vaporize and that will increase (CO2) which makes it shift to the left. Or you can think pCO2 will be increased so pressure shifts it to the side with fewer moles which is to the left also. Nice that both gives the same answer.