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Chapter 12: Problem 100

A sample of pure \(\mathrm{SO}_{3}\) weighing \(0.8312 \mathrm{~g}\) was placed into a 1.00 - \(\mathrm{L}\) flask, sealed, and heated to \(1100 . \mathrm{K}\) to decompose it partially. $$ 2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) $$ If a total pressure of 1.295 atm was developed, calculate the value of \(K_{\mathrm{c}}\) for this reaction at this temperature.

Short Answer

Expert verified
Kc is calculated using equilibrium pressures derived from total pressure.

Step by step solution

01

Find the initial moles of SO3

First, calculate the moles of \( \mathrm{SO}_3 \) using its molar mass. The molar mass of \( \mathrm{SO}_3 \) is 80.07 g/mol. Thus, the moles of \( \mathrm{SO}_3 \) is given by: \( \text{moles of } \mathrm{SO}_3 = \frac{0.8312 \text{ g}}{80.07 \text{ g/mol}} \approx 0.01038 \text{ mol} \).
02

Set up the equation for equilibrium pressures

The reaction, \( 2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \), starts with only \( \mathrm{SO}_3 \). Let the change in moles for the reaction be \( x \). At equilibrium, the moles will be: \( \mathrm{SO}_3: 0.01038 - x \); \( \mathrm{SO}_2: x \); \( \mathrm{O}_2: \frac{x}{2} \). Since \( \mathrm{SO}_2 \) and \( \mathrm{O}_2 \) are formed in a 2:1 ratio, the same ratio applies to their pressures.
03

Calculate the change in number of moles and total moles at equilibrium

Initially, the total moles is \( 0.01038 \text{ mol (all } \mathrm{SO}_3) \). At equilibrium, total moles are \( 0.01038 - x + x/2 + x \), which simplifies to: \( 0.01038 + \frac{x}{2} \).
04

Calculate equilibrium pressures using total pressure

Given that the total pressure is 1.295 atm, use this to find \( x \). The pressure for each gas is proportional to its mole fraction. The sum of these pressures is 1.295 atm.
05

Set up and solve the pressure equilibrium equation

Let \( P_{\mathrm{SO}_2} = \frac{2x}{2x + (0.01038 - x)} \cdot 1.295 \), \( P_{\mathrm{O}_2} = \frac{x}{2x + (0.01038 - x)} \cdot 1.295 \), and \( x + \frac{x}{2} = 1.295 \). Solve for \( x \) to find equilibrium concentrations.
06

Calculate Kc for the reaction

Express \( K_c \) using the equilibrium concentrations: \( K_c = \frac{[P_{\mathrm{SO}_2}]^2 [P_{\mathrm{O}_2}]}{[P_{\mathrm{SO}_3}]^2} \). Substitute the calculated values of pressures into this equation to obtain \( K_c \).

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, denoted as \( K_c \), is a unique value for a chemical reaction at a given temperature. It measures the ratio of the concentrations of products to reactants when the system is at equilibrium. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of the reactants and products.

For a general reaction, \( aA + bB \rightleftharpoons cC + dD \), the expression for \( K_c \) is:
  • \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
Here, the square brackets represent concentrations, and the exponents correspond to the coefficients of the substances in the balanced equation.

In the case of sulfur trioxide (\( \text{SO}_3 \)) decomposing into sulfur dioxide (\( \text{SO}_2 \)) and oxygen (\( \text{O}_2 \)), the equilibrium expression becomes:

\( K_c = \frac{[\text{SO}_2]^2[\text{O}_2]}{[\text{SO}_3]^2} \)

Calculating \( K_c \) helps predict the extent of the reaction and the concentrations of each species at equilibrium.
SO3 Decomposition
In this exercise, the compound sulfur trioxide (\( \text{SO}_3 \)) undergoes decomposition at high temperature, following the reaction:

\( 2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g) \)

Initially, we start with only \( \text{SO}_3 \). When the flask is heated to 1100 K, some of the \( \text{SO}_3 \) decomposes into \( \text{SO}_2 \) and \( \text{O}_2 \).

Key observations about this decomposition:
  • Stoichiometry: The reaction shows that 2 moles of \( \text{SO}_3 \) produce 2 moles of \( \text{SO}_2 \) and 1 mole of \( \text{O}_2 \).
  • Change in moles: If \( x \) is the amount of \( \text{SO}_3 \) decomposed, the equilibrium moles become \( 0.01038 - x \) for \( \text{SO}_3 \), \( x \) for \( \text{SO}_2 \), and \( \frac{x}{2} \) for \( \text{O}_2 \).
  • Pressure relationship: With a total pressure of 1.295 atm, the partial pressures depend on mole fractions and are used to find the equilibrium concentrations.
Understanding the decomposition process helps in predicting how much \( \text{SO}_3 \) will convert under given conditions.
Molar Mass Calculation
Molar mass is a crucial concept for converting between grams and moles. It is defined as the mass of one mole of a substance, and is typically expressed in g/mol.

To calculate the moles of \( \text{SO}_3 \) from a given mass, you need its molar mass, which is the sum of the atomic masses of its constituent elements:
  • Sulfur (S): approximately 32.07 g/mol
  • Oxygen (O): approximately 16.00 g/mol x 3 = 48.00 g/mol
Add them together to find the molar mass of \( \text{SO}_3 \):
  • 32.07 + 48.00 = 80.07 g/mol
Using this, the moles of \( \text{SO}_3 \) from a mass of 0.8312 g is calculated as:

\( \text{moles of } \text{SO}_3 = \frac{0.8312 \text{ g}}{80.07 \text{ g/mol}} \approx 0.01038 \text{ mol} \)

This calculation is essential for progressing through problems involving chemical equilibria and reactions.

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Most popular questions from this chapter

Use the fact that the equilibrium constant \(K_{\mathrm{c}}\) equals the ratio of the forward rate constant divided by the reverse rate constant, together with the Arrhenius equation \(k=A e^{-E_{\mathrm{a}} / R T}\), to show that a catalyst does not affect the value of an equilibrium constant even though the catalyst increases the rates of forward and reverse reactions. Assume that the frequency factors \(A\) for forward and reverse reactions do not change, and that the catalyst lowers the activation barrier for the catalyzed reaction.

Consider the gas-phase reaction of \(\mathrm{N}_{2}+\mathrm{O}_{2}\) to give \(2 \mathrm{NO}\) and the reverse reaction of 2 NO to give \(\mathbf{N}_{2}+\mathrm{O}_{2},\) discussed in Section 12-2e. An equilibrium mixture of \(\mathrm{NO}\), \(\mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) at \(5000 . \mathrm{K}\) that contains equal concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) has a concentration of NO about half as great. Make qualitatively correct plots of the concentrations of reactants and products versus time for these two processes, showing the initial state and the final dynamic equilibrium state. Assume a temperature of \(5000 . \mathrm{K}\). Don't do any calculations-just sketch how you think the plots should look.

Explain in your own words why it was so important to find a highly effective catalyst for the ammonia synthesis reaction before the Haber-Bosch process could become successful.

The equilibrium constant, \(K_{\mathrm{c}}\), for the reaction $$ \mathrm{Br}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{BrF}(\mathrm{g}) $$ is 55.3 . Calculate what the equilibrium concentrations of all these gases are if the initial concentrations of bromine and fluorine were both \(0.220 \mathrm{~mol} / \mathrm{L}\). (Assume constantvolume conditions.)

For each of these chemical reactions, predict whether the equilibrium constant at \(25^{\circ} \mathrm{C}\) is greater than 1 or less than \(1,\) or state that insufficient information is available. Also indicate whether each reaction is product-favored or reactant-favored. $$ \begin{array}{l} \text { (a) } 2 \mathrm{NaCl}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=823 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=-566 \mathrm{~kJ} / \mathrm{mol} \\ \text { (c) } 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \\ \Delta_{\mathrm{r}} H^{\circ}=2045 \mathrm{~kJ} / \mathrm{mol} \end{array} $$
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