Baking soda (sodium bicarbonate) undergoes thermal decomposition as $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

Short Answer

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More \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) will be produced by adding more baking soda to the reaction mixture in an open vessel compared to a closed one.

Step by step solution

01

Identify the Reaction

The reaction that is given is the thermal decomposition of baking soda (sodium bicarbonate - NaHCO3), which is represented by the equation: \[ 2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \] It’s a reaction with two reactants and three products. The sodium bicarbonate is breaking down into sodium carbonate, carbon dioxide, and water.
02

Applying Le Chatelier's Principle for Closed Vessel

In terms of Le Chatelier’s principle, if we add more baking soda (sodium bicarbonate) to the reaction in a closed vessel, the reaction would shift to the right to consume the extra reactant and therefore produce more products - including more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in order to achieve equilibrium. This is because in a closed system, the concentration of all reactants and products can change.
03

Applying Le Chatelier's Principle for Open Vessel

In an open vessel, the gaseous products ( \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ) have the ability to leave the system. When sodium bicarbonate is added, the reaction will shift towards the right. However, as the gases formed can escape, this keeps the reaction from reaching a new equilibrium state, thus continuously driving the reaction to the right to form more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Therefore, even more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) will be produced in an open vessel compared to a closed one.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's principle, named after French chemist Henri Louis Le Chatelier, is a central concept in chemical equilibrium. It articulates how a dynamic equilibrium adjusts to counteract changes to the system's conditions. In simple terms, it states that if an external change is applied to a system at equilibrium, the system will adjust itself in such a way as to partially counteract that change.

For instance, when the concentration of a reactant is increased, the system reacts by consuming the surplus reactant and producing more products, shifting the equilibrium to the right. Conversely, if we take away a product from the system, it will try to produce more of it, again shifting the equilibrium. This principle also applies to changes in temperature and pressure, adjusting the equilibrium to accommodate these changes while maintaining the balance of the chemical reaction.

In the exercise regarding the decomposition of baking soda, adding more solid baking soda to a closed vessel will cause the system to produce more carbon dioxide and water to achieve a new equilibrium. The system responds to this increase in reactant by shifting the reaction to the right, producing more of the gaseous products.
Chemical Equilibrium
Chemical equilibrium refers to the state of a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time, but their presence does not signify the halt of chemical activity. Instead, both reactions are occurring simultaneously at a constant rate.

It's important to understand that equilibrium is a dynamic state. Even if we cannot always observe it, reactant molecules are constantly converting to products and vice versa. In the context of the baking soda decomposition, even though sodium bicarbonate breaks down to sodium carbonate, carbon dioxide, and water, there is a point at which the amount of baking soda decomposing will equal the amount of sodium carbonate, carbon dioxide, and water recombining into baking soda.

A well-known graphic representation of this process is the equilibrium constant expression, where the concentration of products raised to their stoichiometric coefficients is divided by the concentration of reactants raised to their stoichiometric coefficients. It's significant here because it quantitatively describes the position of equilibrium which helps in predicting how the reaction adjusts to changes like adding more reactant.
Reaction Mechanisms
Reaction mechanisms discuss the step-by-step sequence of elementary reactions by which overall chemical change occurs. Each step within a mechanism describes a fundamental chemical process, such as bond breaking, bond forming, or the transfer of an electron, and is characterized by its own rate of reaction.

The detailed paths taken by atoms and molecules as they rearrange themselves from reactants to products are often complex. In the decomposition of baking soda, although the overall reaction appears straightforward, multiple intermediate steps and species might be involved that are not explicitly detailed in the overall equation. Understanding the mechanism helps chemists not just know the order in which bonds break and form but also why adding more reactant might affect the reaction rates and the quantities of products formed.

Examining mechanisms is essential for predicting how changing conditions, such as temperature or the concentration of a reactant, can influence the speed and outcome of a reaction. In educational contexts, mechanisms reinforce the understanding of concepts like Le Chatelier's principle, as mechanisms provide insight into how the microscopic processes respond to macroscopic changes.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) is 0.83 at \(375^{\circ} \mathrm{C} . \mathrm{A}\) 14.6-g sample of ammonia is placed in a 4.00 -L flask and heated to \(375^{\circ} \mathrm{C}\). Calculate the concentrations of all the gases when equilibrium is reached.

When heated, ammonium carbamate decomposes as $$ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ At a certain temperature the equilibrium pressure of the system is 0.318 atm. Calculate \(K_{P}\) for the reaction.

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 0.65 at \(395^{\circ} \mathrm{C}\). $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is \(K_{\mathrm{c}}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

A quantity of 0.20 mole of carbon dioxide was heated at a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Under this condition, the average molar mass of the gases was found to be \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is the \(K_{P}\) for the equilibrium if the total pressure was 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

What do the symbols \(K_{c}\) and \(K_{P}\) represent?

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