Answer to Question #263499 in General Chemistry for Helen

Question

Sodium bicarbonate undergoes thermal decomposition ac-

cording to the reaction

2NaHCO3(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 of argon gas is added.

(b) NaHCO3(s) is added. (c) Mg(ClO4)2(s) is added as a drying

agent to remove H2O. (d) Dry ice is added at constant T.

Answer 1

a) Step 1

0.20 atm added argon

Argon is an inert gas and is not part of the reaction. Therefore, it will not affect the reaction and there will be no shift in the system.

b) Step 2

Removing $H_2O$ by adding $Mg(ClO_4)_2$

​

H2​O is part of the product side. According to the Le Chatelier's principle, decreasing the concentration of the product will shift the reaction to the right favoring its formation.

The product side is now lacking its product which disturbs the equilibrium. The reaction will produce more of the product to compensate its loss.

c)Step 3

Dry ice is added.

Dry ice is solid CO2. However, it readily sublimes from its solid state to gaseous state. The addition of dry ice as it sublimes will increase the concentration of  CO2. This results to the an equilibrium shift to the left favoring the formation of the reactants.

The product side now has excess component. The reaction will use up the excess concentration to balance out both sides to attain equilibrium again.

Result

(a) no reaction shift

(b) no reaction shift

(c) shift to the right

(d) shift to the left