Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)$$ What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system; (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system; (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.

Short Answer

Expert verified
If CO2 is removed, more NaHCO3 will decompose. Adding Na2CO3 or removing NaHCO3 does not affect the equilibrium position.

Step by step solution

01

Analyzing the effect of removing CO2

According to Le Chatelier's principle, if some CO2 is removed, the reaction will shift to the right side to produce more CO2 in order to restore the equilibrium. This happens because the system always counteracts the changes to reach a stable state, and in this case, it needs to replace the CO2 gone after the removal. Hence, more NaHCO3 will decompose.
02

Analyzing the effect of adding Na2CO3

Adding more Na2CO3 does not affect the equilibrium position because solids and liquids do not appear in the equilibrium expression. The equilibrium is determined by the concentrations of gases and aqueous solutions only. Thus, as long as some Na2CO3 is present to maintain an equilibrium, additional amounts will not change the direction of the equilibrium.
03

Analyzing the effect of removing NaHCO3

If some of the solid NaHCO3 is removed, it will not affect the equilibrium position, for the same reason that adding Na2CO3 does not affect it. But, there would be less overall material available for the reaction, which means less possibility to create products.

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Key Concepts

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

Le Chatelier's Principle
When studying chemical reactions, understanding how a system at equilibrium responds to changes is crucial. Le Chatelier's principle is a foundational concept in chemistry that addresses this very question. Put simply, this principle states that when a stress is applied to a chemical system at equilibrium, the system will adjust in a way that counteracts the stress and reestablishes equilibrium. Stresses include changes in concentration, pressure, volume, or temperature.

For example, consider the removal of a product from a reaction at equilibrium. In response, the equilibrium position will shift towards the product side to replenish what was removed, effectively increasing the concentration of the product until the system reaches a new equilibrium state. This shift can be towards the formation of more products, known as a 'right shift', or more reactants, a 'left shift', depending on which side of the equilibrium the change occurs. Understanding Le Chatelier's principle helps predict the direction of the shift in equilibrium in response to various stresses, which is fundamental when manipulating chemical reactions in industrial processes, laboratory experiments, and even within biological systems.
Equilibrium Position
The equilibrium position of a chemical reaction is a description of the relative concentrations of the reactants and products at equilibrium. At this point, the rate of the forward reaction (reactants producing products) equals the rate of the reverse reaction (products decomposing back into reactants). As a result, the concentrations of reactants and products remain constant over time. An important detail to note is that equilibrium does not mean the concentrations are equal, but rather that they are constant.

The equilibrium of a reaction can be represented by an equilibrium constant (K), which is a numerical value that expresses the ratio of product concentrations to reactant concentrations at equilibrium. Changes to this ratio indicate a shift in the equilibrium position. Therefore, if a change is made to a system at equilibrium (for instance, adding or removing a reactant or product), the equilibrium position will shift to restore the ratio defined by the equilibrium constant, as long as the temperature remains constant. Le Chatelier's principle is pivotal in understanding how these shifts in the equilibrium position occur and what the resulting implications are for the reaction under scrutiny.
Sodium Bicarbonate Decomposition
The decomposition of sodium bicarbonate (baking soda), or (aHCO_3), is an endothermic chemical reaction that is often used to demonstrate the application of chemical equilibrium concepts in an educational setting. When heated, sodium bicarbonate decomposes into sodium carbonate (Na_2CO_3), water (H_2O), and carbon dioxide (CO_2) in a reversible reaction.

This process is characterized by a chemical equilibrium that can be influenced by changes such as the removal or addition of CO_2, or changes to the amount of solid reactant or product present. While changes in the concentration of gases affect the equilibrium position, alterations to the amount of solid present (either sodium bicarbonate or sodium carbonate) do not affect the equilibrium, as solids are not included in the equilibrium expression. However, these changes do affect the total quantity of material that can react, which can have an impact on the quantities of product produced. This distinction is vital for students attempting to understand equilibrium concepts and to predict the outcomes of stressing a system – whether in a classroom experiment or an industrial application.

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Most popular questions from this chapter

The following equilibrium constants were determined at \(1123 \mathrm{~K}\) $$\begin{array}{ll}\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) & K_{P}^{\prime}=1.3 \times 10^{14} \\\\\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) & K_{P}^{\prime \prime}=6.0 \times 10^{-3}\end{array}$$ Write the equilibrium constant expression \(K_{P}\), and calculate the equilibrium constant at \(1123 \mathrm{~K}\) for $$\mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$

Write equilibrium constant expressions for \(K_{\mathrm{c}},\) and for \(K_{P}\), if applicable, for the following processes: (a) \(2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) (b) \(3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)\) (c) \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g)\) (d) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) (e) \(\mathrm{HCOOH}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HCOO}^{-}(a q)\) (f) \(2 \mathrm{HgO}(s) \rightleftharpoons 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\)

Based on rate constant considerations, explain why the equilibrium constant depends on temperature.

Consider this equilibrium reaction in a closed container: $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ What will happen if the following occurs? (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 a \(\mathrm{NaOH}\) solution are added to the mixture. (f) A few drops of a \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water). (g) Temperature is increased.

Pure phosgene gas \(\left(\mathrm{COCl}_{2}\right), 3.00 \times 10^{-2} \mathrm{~mol},\) was placed in a 1.50-L container. It was heated to \(800 \mathrm{~K}\), and at equilibrium the pressure of \(\mathrm{CO}\) was found to be 0.497 atm. Calculate the equilibrium constant \(K_{P}\) for the reaction $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g)$$

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