Question

For the following equilibrium system, which of the changes will form more \(\mathrm{CaCO}_{3} ?\) $$ \begin{array}{r} \mathrm{CO}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-113 \mathrm{~kJ} \end{array} $$ (a) Decrease temperature at constant pressure (no phase change). (b) Increase volume at constant temperature. (c) Increase partial pressure of \(\mathrm{CO}_{2}\). (d) Remove one-half of the initial \(\mathrm{CaCO}_{3}\).

Short answer

Option (a) and (c) will form more \( \mathrm{CaCO_3} \).

Step-by-step solution

Understand the Equilibrium Reaction

The given reaction is in equilibrium: \[ \mathrm{CO_2}(g) + \mathrm{Ca(OH)}_2(s) \rightleftharpoons \mathrm{CaCO_3}(s) + \mathrm{H_2O}(l) \] The reaction is exothermic (\( \Delta H^{\circ} = -113 \mathrm{~kJ} \)). It means heat is released when the reaction proceeds from left to right.

Apply Le Chatelier's Principle

Le Chatelier's Principle states that if an external condition is changed, the equilibrium will shift to counteract the change. We will analyze each option under this principle.

Analyze the Effect of Decreasing Temperature

Decreasing temperature favors the exothermic reaction (produces heat). Since our reaction is exothermic, decreasing temperature will shift the equilibrium to the right, forming more \( \mathrm{CaCO_3} \).

Analyze the Effect of Increasing Volume

Increasing volume decreases the pressure. Because there are more gaseous molecules on the left side (\( \mathrm{CO_2}(g) \)) than on the right side (none), the equilibrium will shift to the left, forming less \( \mathrm{CaCO_3} \).

Analyze the Effect of Increasing Partial Pressure of \(\mathrm{CO_2}\)

Increasing the partial pressure of \( \mathrm{CO_2} \) will increase the concentration of \( \mathrm{CO_2} \), shifting the equilibrium to the right and forming more \( \mathrm{CaCO_3} \).

Analyze the Effect of Removing \( \mathrm{CaCO_3} \)

Removing \( \mathrm{CaCO_3} \) will shift the equilibrium to the right to replace the removed \( \mathrm{CaCO_3} \). However, this does not continuously form more \( \mathrm{CaCO_3} \) but merely tries to replace the amount taken.

Conclusion

Based on Le Chatelier's Principle, the changes that will form more \( \mathrm{CaCO_3} \) are: 1. Decreasing Temperature (Option a) 2. Increasing Partial Pressure of \(\mathrm{CO_2}\) (Option c)

Key concepts

Equilibrium Shift

In chemical reactions, the concept of equilibrium shift is crucial to understand how different conditions affect the reaction balance. According to Le Chatelier's principle, when a change is imposed on a system at equilibrium, the system will adjust itself to counteract that change. For example, in the reaction $$ \text{CO}_2(g) + \text{Ca(OH)}_2(s) \rightleftharpoons \text{CaCO}_3(s) + \text{H}_2O(l) $$ if you decrease the temperature, the equilibrium will shift to the right to produce more heat (since the reaction is exothermic). If you increase the volume, the equilibrium will shift to where there are more gas molecules, which, in this case, is to the left side (producing fewer solid products). Understanding equilibrium shifts can help predict how changes in conditions will affect the quantities of reactants and products in a chemical reaction.

Exothermic Reaction

An exothermic reaction is one that releases heat to its surroundings. In the given equation, $$ \text{CO}_2(g) + \text{Ca(OH)}_2(s) \rightleftharpoons \text{CaCO}_3(s) + \text{H}_2O(l), \text{with} \text{ } \text{ } \triangle H^{\text{o}} = -113 \text{ kJ} $$ we see that the reaction releasing -113 kJ of energy means it's exothermic. This is important when considering the impact of temperature changes. When temperature drops, exothermic reactions produce more products to release heat and counteract the temperature drop. Conversely, if we increase temperature, the reaction will shift to the left, reducing heat production.

Partial Pressure Impact

Partial pressure refers to the pressure exerted by a particular gas in a mixture of gases. In our reaction, the partial pressure of $$ \text{CO}_2(g) $$ can significantly impact the equilibrium. According to Le Chatelier's Principle, increasing the partial pressure of a reactant will shift the equilibrium towards the product side. If we increase the partial pressure of $$ \text{CO}_2 $$ in the reaction $$ \text{CO}_2(g) + \text{Ca(OH)}_2(s) \rightleftharpoons \text{CaCO}_3(s) + \text{H}_2O(l) $$ it increases the concentration of $$ \text{CO}_2 $$ , causing the equilibrium to shift to the right and form more $$ \text{CaCO}_3 $$ . This way, we can control how much product we form by manipulating gas pressure.

Temperature Effect

Temperature can greatly influence the direction of an equilibrium reaction, especially an exothermic one. For the exothermic reaction, $$ \text{CO}_2(g) + \text{Ca(OH)}_2(s) \rightleftharpoons \text{CaCO}_3(s) + \text{H}_2O(l) $$ with $$ \triangle H^{\text{o}} = -113 \text{ kJ} $$ , a decrease in temperature favors the formation of more $$ \text{CaCO}_3 $$ . This happens because the reaction releases heat and lowering the temperature drives the equilibrium to produce more heat, hence more product. On the other hand, increasing temperature would supply additional heat, thus driving the reaction backward to consume some of the added heat, resulting in less product. This dynamic helps us understand how crucial temperature manipulation is in chemical manufacturing and laboratory settings.