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Chapter 12: Problem 64

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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Most popular questions from this chapter

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.

WEB Nitrogen dioxide can decompose to nitrogen oxide and oxygen. $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ \(K\) is \(0.87\) at a certain temperature. A 5.0-L flask at equilibrium is determined to have a total pressure of \(1.25\) atm and oxygen to have a partial pressure of \(0.515\) atm. Calculate \(P_{\mathrm{NO}}\) and \(P_{\mathrm{NO}}\), at equilibrium.

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}} \text { (atm) } & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{B} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\\\\hline\end{array}$$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after 30 s? After 90 s?

Iodine chloride decomposes at high temperatures to iodine and chlorine gases. $$2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g)$$ Equilibrium is established at a certain temperature when the partial pressures of ICI, \(\mathrm{I}_{2}\), and \(\mathrm{Cl}_{2}\) are (in atm) \(0.43,0.16\), and \(0.27\), respectively. (a) Calculate \(K\). (b) If enough iodine condenses to decrease its partial pressure to \(0.10 \mathrm{~atm}\), in which direction will the reaction proceed? What is the partial pressure of iodine when equilibrium is re-established?

Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?
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