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Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

Short Answer

Expert verified
The reaction shifts to the right when the volume is increased, \(\mathrm{CaCO}_3\) is removed, \(\mathrm{NaOH}\) is added, or the temperature is increased (assuming endothermic reaction). On the contrary, it shifts to the left when \(\mathrm{CaO}\) or \(\mathrm{CO}_2\) are added, or \(\mathrm{HCl}\) is added.

Step by step solution

01

Effect of Changing Volume

In the selected reaction, volume change will only affect gases. According to Le Chatelier's principle, the side with more gas molecules is favored when the volume increases. As there is only one gas molecule (\(CO_2\)) the reaction would go to the right to compensate the volume increase.
02

Effect of Adding or Removing Substances

Adding \(\mathrm{CaO}\) (part b) or \(\mathrm{CO}_2\) (part d) would shift the reaction to the left, as the system would strive to achieve equilibrium by reducing the excess substance concentration. Conversely, removing \(\mathrm{CaCO}_3\) (part c) would shift the reaction to the right to replace the \(\mathrm{CaCO}_3\) that was removed.
03

Effect of Adding NaOH or HCl

Adding \(\mathrm{NaOH}\) (part e) or \(\mathrm{HCl}\) (part f) would lead to their respective reactions with \(\mathrm{CaO}\) or \(\mathrm{CaCO}_3\), reducing their concentrations and shifting the original reaction accordingly. The reaction shifts to the right when \(\mathrm{NaOH}\) is added, as \(\mathrm{CaO}\) is consumed. It shifts to the left when \(\mathrm{HCl}\) is added, as \(\mathrm{CaCO}_3\) is consumed.
04

Analysis of Temperature Increase

Increasing the temperature (part g) affects the reaction direction based on its reaction type (exothermic or endothermic). Assuming that this reaction is endothermic (since the formation of \(\mathrm{CaO}\) and \(\mathrm{CO}_2\) usually requires heat), an increase in temperature would shift the reaction to the right, as the system endeavors to lower the temperature by absorbing the additional heat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Understanding the behavior of chemical reactions under stress is central to predicting how the system will respond. Le Chatelier's Principle is a vital concept in chemistry that helps us understand this behavior. It states that when a dynamic equilibrium system is subjected to an external change (such as concentration, temperature, or pressure), the system will adjust itself to minimize the impact of this change. This is much like a balancing act where the system tries to maintain its equilibrium state by shifting the reaction in the direction that will oppose the change.
Reaction Direction Shift
When we apply Le Chatelier's Principle to a reaction, we can determine which way a reaction will shift to re-establish equilibrium. If a reactant or product is added to the system, the reaction shifts away from the increased substance. Conversely, if a reactant or product is removed, the reaction shifts towards the side where the substance was diminished. This shift helps to balance the concentrations and stabilize the system under new conditions. It's an automatic, self-regulatory response similar to how our body temperature adjusts to external temperature changes.
Effects of Volume Change in Gases
Changes in volume notably impact reactions involving gases. As a rule of thumb, decreasing the volume increases pressure, thus favoring the side of the reaction with fewer gas molecules. Conversely, increasing the volume decreases pressure and favors the side with more gas molecules. This can be visualized by imagining a crowded room where people represent gas molecules; if the room size expands (volume increases), there is more space for 'molecules' to spread out comfortably.
Impact of Temperature on Equilibrium
Temperature changes can dramatically influence the direction of an equilibrium reaction. Heating typically favors endothermic reactions, where the system absorbs heat. Cooling, on the other hand, favors exothermic reactions, which release heat. This is because, according to Le Chatelier’s Principle, the system will adjust to absorb extra heat or release some when it's taken away. It's similar to how we put on a sweater when it's cold (to add heat) or take it off when it's hot (to lose heat).
Equilibrium and Concentration Changes
Concentration changes in a chemical equilibrium can cause the reaction to shift in such a way as to counteract the change. If the concentration of a reactant is increased, the reaction tends to produce more product. If the concentration of a product is increased, the reaction will shift to produce more reactants. This is akin to adding or taking away game pieces from one side of a scale to keep it balanced. Keeping an eye on concentrations is crucial in industrial reactions where optimal yields are desired.

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Most popular questions from this chapter

A sealed glass bulb contains a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases. When the bulb is heated from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\), what happens to these properties of the gases: (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) ), (e) density? Assume that volume remains constant. (Hint: \(\mathrm{NO}_{2}\) is a brown gas; \(\mathrm{N}_{2} \mathrm{O}_{4}\) is colorless.)

Define equilibrium. Give two examples of a dynamic equilibrium.

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C} . \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are mixed initially at 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures, 1.112 atm?

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00 -L flask. At \(648 \mathrm{~K}\), there is 0.0345 mole of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$

Consider the potential energy diagrams for two types of reactions \(\mathrm{A} \rightleftharpoons \mathrm{B}\). In each case, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture.

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