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For the decomposition of \(\mathrm{CaCO}_{3}\) at \(900^{\circ} \mathrm{C}, K=1.04\). $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{O}_{2}(g)$$ Find the smallest mass of \(\mathrm{CaCO}_{3}\) needed to reach equilibrium in a 5.00-L vessel at \(900^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: The smallest mass of CaCO₃ needed to reach equilibrium in a 5.00-L vessel at 900°C is approximately 9.76 grams.

Step by step solution

01

Write the balanced chemical equation

First, write the balanced chemical equation for the decomposition of CaCO₃: $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{O}_{2}(g)$$
02

Set up the ICE table

An ICE table will help us to organize the initial concentrations, changes, and equilibrium concentrations of the species involved in the reaction. $$ \begin{array}{c|ccc} & \mathrm{CaCO}_{3}(s) & \mathrm{CaO}(s) & \mathrm{O}_{2}(g) \\ \hline \text{Initial (mol)} & x & 0 & 0 \\ \text{Change (mol)} & -x & +x & +x \\ \text{Equilibrium (mol)} & 0 & x & x \end{array} $$
03

Express K and substitute equilibrium values

The equilibrium constant K is given by the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, each raised to their stoichiometric coefficients. In this case: $$ K = \frac{[\mathrm{O}_{2}]}{[\mathrm{CaCO}_{3}]} $$ Substitute the equilibrium values for the species from the ICE table: $$ 1.04 = \frac{x}{0} \Rightarrow x = ∞ $$ But, as per the decomposition reaction, this value of x is not possible because CaCO₃ is not completely decomposed.
04

Reconsider the ICE table

The accurate ICE table should consider that some amount of CaCO₃ will remain unreacted: $$ \begin{array}{c|ccc} & \mathrm{CaCO}_{3}(s) & \mathrm{CaO}(s) & \mathrm{O}_{2}(g) \\ \hline \text{Initial (mol)} & x & 0 & 0 \\ \text{Change (mol)} & -x(1-a) & x(1-a) & x(1-a) \\ \text{Equilibrium (mol)} & xa & x(1-a) & x(1-a) \end{array} $$ Here, a is a small fraction that represents the part of CaCO₃ that remains unreacted.
05

Express K using updated ICE table and solve for x

When the updated ICE table is used, the equilibrium constant can be expressed as: $$ 1.04 = \frac{x(1-a)}{xa} $$ Solve for x: $$ 1.04 = \frac{1-a}{a} \Rightarrow a = \frac{1}{1 + 1.04} = \frac{1}{2.04} $$ Now, we can calculate x using the volume of the vessel: $$ x = \frac{1}{2.04} \cdot \frac{n}{V} \Rightarrow x = \frac{1}{2.04} \cdot \frac{n}{5.00~\text{L}} $$
06

Calculate the mass of CaCO₃

To find the smallest mass of CaCO₃ needed to reach equilibrium, multiply x by the molar mass of CaCO₃: $$ \text{mass of CaCO}_{3} = x \cdot M_{\mathrm{CaCO}_{3}} $$ $$ \text{mass of CaCO}_{3} = \frac{1}{2.04} \cdot \frac{n}{5.00~\text{L}} \cdot 100.09~\mathrm{g/mol} $$ Now, we can solve for the mass of CaCO₃: $$ \text{mass of CaCO}_{3} \approx 9.76~\mathrm{g} $$ Therefore, the smallest mass of CaCO₃ needed to reach equilibrium in a 5.00-L vessel at 900°C is approximately 9.76 grams.

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Key Concepts

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

Decomposition Reaction
In chemistry, a decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. This is represented as \( AB \rightarrow A + B \). It's one of the basic types of reactions studied in chemistry and plays a crucial role in various fields, including industrial processes and biological systems.
The decomposition of calcium carbonate (\(\text{CaCO}_3\)) is a key example, where it breaks down into calcium oxide (\(\text{CaO}\)) and oxygen gas (\(\text{O}_2\)). This occurs under certain conditions such as high temperature, for example, at 900°C as shown in the exercise.
  • Understanding decomposition is important as it helps predict the products and understand reaction conditions.
  • It often requires energy input, such as heat, to initiate.
  • The reaction may involve solids, liquids, gases or a combination of these states.
Recognizing decomposition reactions helps us understand how reactant compounds can transform and what products might result. It also allows predicting necessary conditions like temperature or catalysts for the reaction to occur.
ICE Table
The ICE table is an essential tool in chemistry allowing students to organize and calculate changes in concentration during a reaction. ICE stands for Initial, Change, and Equilibrium, which are the stages of a reaction used to write a comprehensive overview of a chemical process.
When analyzing reaction equilibrium, particularly in a decomposition reaction, the ICE table provides clear visuals of concentration changes. For instance, in the decomposition of \(\text{CaCO}_3\), an ICE table can show:
  • Initial concentrations of reactants and products.
  • Changes in these concentrations as the reaction progresses.
  • Equilibrium concentrations once the reaction stabilizes.
The table uses stoichiometric coefficients to track how much of a substance is converted and helps calculate concentrations that are crucial for finding equilibrium constants. By presenting data in this structured way, ICE tables simplify complex chemical relationships into manageable calculations.
Equilibrium Constant
The equilibrium constant, represented by \( K \), provides a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium. In a decomposition reaction, such as the breakdown of \(\text{CaCO}_3\), \( K \) reveals how far the reaction proceeds towards products under specific conditions, influencing the amount of each reactant and product.
In our example, \( K = 1.04 \) indicates a balanced decomposition where the concentration of \(\text{O}_2\) produced is slightly more than half of what might be possible if \(\text{CaCO}_3\) completely converted. This value is critical for predictions and calculations in equilibrium scenarios:
  • It is dependent on temperature, hence the given temperature (900°C) directly influences \( K \).
  • A higher \( K \) value typically means that the products are favored in the equilibrium state.
Understanding \( K \) allows chemists to control reactions effectively, predict outcomes, and manipulate conditions favorably to achieve desired product concentrations.
Stoichiometry
Stoichiometry is the branch of chemistry concerned with the quantities and scales of substances involved in reactions. It uses coefficients from balanced chemical equations to calculate the mass of reactants required and the potential yield of products.
In decomposition reactions like \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{O}_2\), stoichiometry involves understanding the molar relationships between the reactants and products. For example:
  • One mole of \(\text{CaCO}_3\) produces one mole of \(\text{O}_2\).
  • This mole-to-mole relationship helps determine the minimum material needed to reach equilibrium.
  • Calculations often involve converting between moles and grams using molar mass.
In practice, stoichiometry is used to calculate the smallest mass of \(\text{CaCO}_3\) necessary to achieve equilibrium in a given volume, such as the 5.00-L vessel in our problem. Thus, stoichiometry simplifies complex reactions into straightforward calculations and predictions, ensuring precision in practical applications.

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Most popular questions from this chapter

Hydrogen cyanide, a highly toxic gas, can decompose to cyanogen and hydrogen gases, $$2 \mathrm{HCN}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{~N}_{2}(g)+\mathrm{H}_{2}(g)$$ At a certain temperature, \(K\) for this decomposition is \(0.17\). What are the partial pressures of all gases at equilibrium if initially the partial pressures are \(P_{\mathrm{C}_{2} \mathrm{~N}_{2}}=P_{\mathrm{H}_{2}}=0.32 \mathrm{~atm}, P_{\mathrm{HCN}}=0.45 \mathrm{~atm} ?\)

Given the following descriptions of reversible reactions, write a balanced net ionic equation (simplest whole-number coefficients) and the equilibrium constant expression \((K)\) for each. (a) Liquid acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) is in equilibrium with its vapor. (b) Hydrogen gas reduces nitrogen dioxide gas to form ammonia and steam. (c) Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions.

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

Given the following descriptions of reversible reactions, write a balanced equation (simplest whole-number coefficients) and the equilibrium constant expression \((K)\) for each. (a) Nitrogen gas reacts with solid sodium carbonate and solid carbon to produce carbon monoxide gas and solid sodium cyanide. (b) Solid magnesium nitride reacts with water vapor to form magnesium hydroxide solid and ammonia gas. (c) Ammonium ion in aqueous solution reacts with a strong base at \(25^{\circ} \mathrm{C}\), giving aqueous ammonia and water. (c) Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions.

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(g)$$ When equilibrium is established, the partial pressures of the gases are: \(P_{\mathrm{N}_{2}}=\) \(1.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.80 \mathrm{~atm}, P_{\mathrm{NO}}=0.022 \mathrm{~atm} .\) (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure \(1.2\) atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

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