Problem 39 - Chemistry and Chemical Reactivity

Question

Heating a metal carbonate leads to decomposition. $$ \mathrm{BaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ Predict the effect on the equilibrium of each change listed below. Answer by choosing (i) no change, (ii) shifts left, or (iii) shifts right. (a) add \(\mathrm{BaCO}_{3}\) (c) add BaO (b) add \(\mathrm{CO}_{2}\) (d) raise the temperature (e) increase the volume of the flask containing the reaction

Step-by-Step Solution

Answer

(a) No change, (c) No change, (b) Shifts left, (d) Shifts right, (e) Shifts right.

Step 1: Analyze the Reaction

The decomposition reaction of barium carbonate \(\mathrm{BaCO}_3(\mathrm{s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})\) is an equilibrium reaction involving both solid and gaseous species. According to Le Châtelier's principle, any change in concentration, pressure, volume, or temperature of a reaction will result in a shift of the equilibrium to oppose the change.

Step 2: Effect of Adding \(\mathrm{BaCO}_3\)

Since \(\mathrm{BaCO}_3\) is a solid, adding more \(\mathrm{BaCO}_3\) does not affect the equilibrium position because solids are not included in the equilibrium expression. The equilibrium will experience no shift.****Conclusion:**** No change (i).

Step 3: Effect of Adding \(\mathrm{BaO}\)

Similarly, \(\mathrm{BaO}\) is also a solid and adding it will not affect the equilibrium position for the same reasons as \(\mathrm{BaCO}_3\). Solids do not appear in the equilibrium constant expression.****Conclusion:**** No change (i).

Step 4: Effect of Adding \(\mathrm{CO}_2\)

Adding more \(\mathrm{CO}_2\) increases the concentration of the gaseous reactant product. According to Le Châtelier's principle, the system will attempt to counteract this change by shifting the equilibrium to the left, favoring the formation of \(\mathrm{BaCO}_3\).****Conclusion:**** Shifts left (ii).

Step 5: Effect of Raising the Temperature

The reaction is endothermic when considering the decomposition of \(\mathrm{BaCO}_3\) because breaking bonds typically requires energy. Increasing the temperature provides additional energy, favoring the endothermic direction, which in this case is the decomposition. Thus, the equilibrium shifts to the right to absorb the added energy.****Conclusion:**** Shifts right (iii).

Step 6: Effect of Increasing the Volume of the Flask

Increasing the volume of the flask decreases the pressure of the system. Since there is a greater number of moles of gas on the right side of the equation, the equilibrium will shift to the right to increase the pressure by producing more \(\mathrm{CO}_2\) gas.****Conclusion:**** Shifts right (iii).

Key Concepts

Chemical Equilibrium

In any reversible chemical process, a state of chemical equilibrium is achieved when the rates of the forward and reverse reactions become equal. At this point, the concentrations of reactants and products remain constant over time. The equilibrium state does not mean that the reactants and products are equal in concentration; it only means that their rates of formation are balanced.

Le Châtelier's Principle is a fundamental chemical principle that helps predict the behavior of a reaction in equilibrium when it is subjected to an external change. This can include changes in concentration, temperature, or pressure. The system will seek to minimize the effect of the change by shifting the position of the equilibrium to counteract it.

For instance, in the case of adding or removing reactants or products, the equilibrium will shift towards the side that helps restore balance. If a reactant is added, the system will favor the formation of the product, and vice versa, until a new equilibrium is established.

Endothermic Reaction

An endothermic reaction is a chemical reaction that absorbs energy from its surroundings, typically in the form of heat. In these reactions, the energy required to break bonds in the reactants is greater than the energy released when new bonds are formed in the products.

In the decomposition reaction of $ BaCO_3$ to form $ BaO$ and $ CO_2$, the process is endothermic. This means upon increasing the temperature, the equilibrium will shift towards the products side to absorb the additional heat energy.

This is a practical illustration of Le Châtelier's principle, where increasing the energy available to the system encourages the reaction to proceed in the direction that "uses up" that energy—namely, the endothermic breakdown of $ BaCO_3$.

Pressure-Volume Relationship

The pressure-volume relationship in a gaseous system at equilibrium is another crucial aspect of Le Châtelier's principle. Essentially, this principle states that a change in volume will be counteracted by a shift in the equilibrium to restore pressure, according to the ideal gas law ($PV = nRT$).

In our reaction, increasing the volume of the flask is equivalent to decreasing the pressure, as the same number of gas molecules now occupy a larger space. To combat this decrease in pressure, the system will shift to the side of the reaction that contains more gas molecules.

Since the right side of the reaction $ (BaO + CO_2)$ has more moles of gas ( $ CO_2$), the equilibrium will shift to the right, increasing the production of gas to restore lost pressure.

Reaction Shift Analysis

Reaction shift analysis involves predicting the direction in which the equilibrium will move in response to changes in the system.

When additional $ CO_2$ is introduced into the reaction mixture, the balance is tilted by increasing the concentration of gaseous product. According to Le Châtelier's principle, this imbalance prompts a shift towards the reactant side $ (BaCO_3)$ to counteract the rise in $ CO_2$, reducing its concentration back to a balanced state.

On the other hand, when the temperature is increased, the reaction favors the endothermic decomposition (rightward shift), using this added energy to form more $ BaO$ and $ CO_2$. Meanwhile, increasing the flask volume shifts the equilibrium to the right as a response to decreased pressure, also favoring the formation of $ BaO$ and $ CO_2$. In analyzing these shifts, the core idea is always the same: the equilibrium will move to oppose the imposed change, seeking a new balance point under the altered conditions.