Question

1. consider the following equilibrium: $caco_3(s)

ightleftharpoons ca^{2 + }(aq)+co_3^{2 - }(aq) delta h>0$. in which direction will the equilibrium shift if a. $na_2co_3$ is added? why? b. $cacl_2$ is added? why? c. $caco_3$ is added? why? d. the temperature is increased? why? e. hcl is added? why?

Explanation:

Step1: Analyze adding $Na_2CO_3$

Adding $Na_2CO_3$ increases the concentration of $CO_3^{2 - }(aq)$. According to Le - Chatelier's principle, the system will shift to counteract the change. So the equilibrium will shift to the left to reduce the increased $CO_3^{2 - }$ concentration.

Step2: Analyze adding $CaCl_2$

Adding $CaCl_2$ increases the concentration of $Ca^{2 + }(aq)$. By Le - Chatelier's principle, the equilibrium will shift to the left to decrease the added $Ca^{2 + }$ concentration.

Step3: Analyze adding $CaCO_3$

$CaCO_3$ is a solid. Changing the amount of a solid in a heterogeneous equilibrium does not affect the equilibrium position as the concentration of a pure solid is considered constant. So the equilibrium does not shift.

Step4: Analyze increasing temperature

The reaction is endothermic ($\Delta H>0$). Increasing the temperature favors the endothermic direction. So the equilibrium will shift to the right to absorb the added heat.

Step5: Analyze adding $HCl$

$HCl$ will react with $CO_3^{2 - }(aq)$: $2H^+(aq)+CO_3^{2 - }(aq)=H_2O(l) + CO_2(g)$. This decreases the concentration of $CO_3^{2 - }(aq)$. According to Le - Chatelier's principle, the equilibrium will shift to the right to replenish the decreased $CO_3^{2 - }$ concentration.

Answer:

A. The equilibrium will shift to the left because the concentration of $CO_3^{2 - }$ increases.
B. The equilibrium will shift to the left because the concentration of $Ca^{2 + }$ increases.
C. The equilibrium does not shift because $CaCO_3$ is a solid and its concentration is constant.
D. The equilibrium will shift to the right because the reaction is endothermic and increasing temperature favors the endothermic direction.
E. The equilibrium will shift to the right because the concentration of $CO_3^{2 - }$ decreases due to its reaction with $HCl$.