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Chapter 15: Problem 55

Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftarrows \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g) $$ 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
(a) Shift right, (b) No change, (c) No change.

Step by step solution

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01

Understanding Equilibrium Shift

According to Le Chatelier's Principle, if an external change is applied to a system at equilibrium, the system adjusts to counteract that change.
02

Analyzing Removal of CO2

When some \( ext{CO}_2\) is removed, the system will try to counter this change by producing more \( ext{CO}_2\). The equilibrium will shift to the right, favoring the production of \( ext{Na}_2 ext{CO}_3, ext{H}_2 ext{O},\) and \( ext{CO}_2\).
03

Analyzing Addition of Na2CO3

Adding more solid \( ext{Na}_2 ext{CO}_3 \) will not affect the equilibrium position because it is a pure solid, which does not appear in the equilibrium expression. The concentrations of gases, however, remain unchanged and thus the equilibrium remains unshifted.
04

Analyzing Removal of NaHCO3

Removing some solid \( ext{NaHCO}_3 \) will also not affect the equilibrium position, as it is a pure solid and does not appear in the equilibrium expression. The equilibrium remains unshifted.

Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemical equilibrium. It explains how a system at equilibrium responds to external changes. When a system in equilibrium undergoes a change in concentration, pressure, or temperature, the system adjusts to minimize that change and restore a new equilibrium.

Imagine a balanced seesaw. If you add weight to one side, the seesaw will tip, and the system seeks to counterbalance and stabilize once more. Similarly, in a chemical system, if concentrations or other conditions are altered, the reaction will shift to restore equilibrium.
  • If a reactant is added, the equilibrium shifts towards the products.
  • If a product is removed, the equilibrium shifts towards the side of the removed product to replace it.
  • Pressure changes can also affect equilibrium, especially in reactions involving gases.
Understanding Le Chatelier's Principle helps predict the outcomes of chemical reactions under varying conditions and is crucial for industries that rely on chemical manufacturing.
Sodium Bicarbonate Reaction
The reaction of sodium bicarbonate (\( ext{NaHCO}_3 \) ) when heated is a classic example of a decomposition reaction. In a closed system, solid sodium bicarbonate decomposes into sodium carbonate (\( ext{Na}_2 ext{CO}_3 \) ), water vapor (\( ext{H}_2 ext{O} \) ), and carbon dioxide gas (\( ext{CO}_2 \) ). The equation is as follows:
\[2 ext{NaHCO}_3(s) ightleftarrows ext{Na}_2 ext{CO}_3(s) + ext{H}_2 ext{O}(g) + ext{CO}_2(g)\]

This reaction is interesting because it involves both solid and gaseous products. The gases play a significant role in altering the equilibrium because they can easily be affected by changes in pressure and concentration.

It's essential to note that while solids like sodium carbonate and sodium bicarbonate appear in the reaction, their quantities don't directly affect the equilibrium position. Only the concentrations of gaseous components, such as \( ext{H}_2 ext{O} \) and \( ext{CO}_2 \), shift the equilibrium according to Le Chatelier's Principle.
Equilibrium Shift
The concept of equilibrium shift refers to the movement of a chemical reaction's equilibrium position in response to a change in conditions. When analyzing equilibrium shifts, it's crucial to think about what causes the shift and the direction it moves.

In the given exercise:
  • When \( ext{CO}_2 \) is removed, Le Chatelier's Principle predicts the system will compensate by shifting the equilibrium towards the right to produce more \( ext{CO}_2 \).
  • Adding more solid \( ext{Na}_2 ext{CO}_3 \) does not change the equilibrium because solids don't affect equilibrium concentrations.
  • Similarly, removing solid \( ext{NaHCO}_3 \) does not shift the equilibrium for the same reason.
In summary, only changes in conditions that affect gas concentrations or external pressure will typically alter a closed system's equilibrium position. Understanding equilibrium shifts helps predict how reactions respond to various interventions, which is essential in both academic studies and practical applications.
Closed System Reactions
A closed system in chemistry refers to a scenario where a reaction takes place in a container that does not allow exchange of matter with the surroundings. However, energy can still be transferred, typically in the form of heat.

In the sodium bicarbonate reaction, the closed system ensures that gases like \( ext{CO}_2 \) and \( ext{H}_2 ext{O} \) produced cannot escape. This containment allows the gas concentrations to build up, affecting the equilibrium position.
  • Closed systems are vital for accurately studying equilibrium because they maintain consistent matter balance throughout the reaction.
  • Any changes made within a closed system directly impact the internal reaction dynamics.
Studying reactions in closed systems allows scientists to observe true equilibrium behavior without the interference of outside factors, making it easier to apply Le Chatelier's Principle and predict equilibrium shifts.

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Most popular questions from this chapter

One mole of \(\mathrm{N}_{2}\) and three moles of \(\mathrm{H}_{2}\) are placed in a flask at \(375^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is 0.21 . The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}\).

What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant, and, in each case, the reactants are in a cylinder fitted with a movable piston. (a) \(\mathrm{A}(s) \rightleftarrows 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftarrows \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftarrows \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftarrows \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g)\)

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{c}\) and \(K_{p}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=206 \mathrm{~kJ} / \mathrm{mol}\) The secondary stage is carried out at about \(1000^{\circ} \mathrm{C},\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftarrows \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\) (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

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