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Baking soda (sodium bicarbonate) undergoes thermal decomposition as follows: $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftarrows \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ 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

Expert verified
More \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are obtained in an open vessel.

Step by step solution

01

Understanding the Reaction

The reaction in question is the thermal decomposition of baking soda, which is: \[ 2 \mathrm{NaHCO}_{3}(s) \rightleftarrows \mathrm{Na}_{2}\mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \] This reaction is reversible, and it involves solid sodium bicarbonate (\mathrm{NaHCO}_{3}) decomposing into sodium carbonate (\mathrm{Na}_{2} \mathrm{CO}_{3}), carbon dioxide (\mathrm{CO}_{2}), and water vapor (\mathrm{H}_{2} \mathrm{O}).
02

Applying Le Chatelier's Principle

To determine whether the addition of more baking soda will yield more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\), consider Le Chatelier's Principle. Increasing the concentration of a reactant typically shifts the equilibrium towards the products in order to minimize the change, which would increase the amount of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced.
03

Case Analysis - Closed Vessel

In a closed vessel, the system is sealed, meaning that the gaseous products (\(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\)) cannot escape. As \(\mathrm{NaHCO}_{3}\) is added, the reaction will shift to the right to produce more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). However, as these gases build up, equilibrium is reached again without releasing excess gas, which can limit their maximum concentration.
04

Case Analysis - Open Vessel

In an open vessel, gases like \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) can escape the system as they are formed. Adding more \(\mathrm{NaHCO}_{3}\) will continuously shift the reaction towards the products, increasing \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) production as these gases escape, preventing the reaction from reaching equilibrium and allowing more gas to be continually produced.
05

Conclusion

Comparing the two scenarios, more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are produced in an open vessel because the gases can escape, allowing the reaction to keep shifting right without limitation.

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Key Concepts

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

Sodium Bicarbonate Decomposition
In chemistry, sodium bicarbonate, or baking soda, is a common compound with various uses, including in cooking and as a cleaning agent. A fascinating phenomenon occurs when sodium bicarbonate is subjected to heat; it undergoes thermal decomposition. This process involves the breakdown of sodium bicarbonate (\(\mathrm{NaHCO}_{3}\)) into three simpler products: sodium carbonate (\(\mathrm{Na}_{2}\mathrm{CO}_{3}\)), carbon dioxide (\(\mathrm{CO}_{2}\)), and water vapor (\(\mathrm{H}_{2}\mathrm{O}\)).This chemical reaction is important because it is a reversible reaction, meaning it can proceed in both the forward and backward directions. When heated, the forward reaction dominates, producing the solid and gaseous products. The reaction can be represented by the equation:\[2 \mathrm{NaHCO}_{3}(s) \rightleftarrows \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\]Understanding this decomposition is crucial for applications like baking, where the release of carbon dioxide helps dough rise. Thermal decomposition is not only useful in baking but also provides insights into chemical behavior and reaction dynamics.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that helps us predict the effect of changes on a chemical system at equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and re-establish equilibrium.In the context of sodium bicarbonate decomposition, if extra baking soda is added to the mixture, Le Chatelier's Principle can be applied. Adding more sodium bicarbonate increases the concentration of reactants and disturbs the equilibrium. According to the principle, the equilibrium will shift towards the production of more products (\(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\)) to balance the change.This shift in equilibrium is particularly evident in closed and open systems:
  • In a closed system, gases cannot escape, so the shift eventually reaches a new equilibrium.
  • In an open system, gases like \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) can escape, continually driving the reaction forward.
Understanding Le Chatelier's Principle allows chemists and students to predict and control the outcomes of reactions, making it a powerful tool in both academic and practical settings.
Equilibrium Reactions
Equilibrium reactions are a fundamental aspect of chemistry, where the rate of the forward reaction equals the rate of the reverse reaction, resulting in a stable balance between products and reactants. In the decomposition of sodium bicarbonate, equilibrium is achieved when the decomposition rate into sodium carbonate, carbon dioxide, and water vapor matches their recombination.The concept of equilibrium is dynamic rather than static, meaning that reactions continue to occur, but their effects cancel out, maintaining a stable concentration of substances. Several factors can affect equilibrium, including pressure, temperature, and concentration.In the given reaction, the presence of gaseous products (\(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\)) allows us to explore how changes affect equilibrium:
  • Increasing pressure or temperature can shift equilibrium.
  • Removing products (especially gases) in an open system shifts equilibrium to favor product formation continuously.
Understanding the nature of equilibrium reactions helps in manipulating chemical processes, making it essential knowledge in chemistry. Mastery of these concepts enables the prediction and optimization of chemical yields in both laboratory and industrial scenarios.

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Most popular questions from this chapter

When a gas was heated under atmospheric conditions, its color deepened. Heating above \(150^{\circ} \mathrm{C}\) caused the color to fade, and at \(550^{\circ} \mathrm{C}\) the color was barely detectable. However, at \(550^{\circ} \mathrm{C},\) the color was partially restored by increasing the pressure of the system. Which of the following best fits the preceding description: (a) a mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color, and nitrogen dioxide is a brown gas. The other gases are colorless.) Justify your choice.

Does the addition of a catalyst have any effects on the position of an equilibrium?

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

Water is a very weak electrolyte that undergoes the following ionization (called autoionization): $$ \mathrm{H}_{2} \mathrm{O}(l) \stackrel{k_{1}}{\stackrel{\mathrm{m}_{-1}}} \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (a) If \(k_{1}=2.4 \times 10^{-5} \mathrm{~s}^{-1}\) and \(k_{-1}=1.3 \times 10^{11} / M \cdot \mathrm{s}\) calculate the equilibrium constant \(K\) where \(K=\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right] /\left[\mathrm{H}_{2} \mathrm{O}\right] .\) (b) Calculate the product \(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right],\) \(\left[\mathrm{H}^{+}\right],\) and \(\left[\mathrm{OH}^{-}\right] .\) (Hint : Calculate the concentration of liquid water using its density, \(1.0 \mathrm{~g} / \mathrm{mL}\).)

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}:\) $$\mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftarrows \mathrm{NaCl}(l)+\mathrm{K}(g)$$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

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