Question

A 5.37 -g sample of \(\mathrm{SO}_{3}\) is placed in a 5.00-L container and heated to \(1000 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the total pressure in the container is 157 \(\mathrm{kPa}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Short answer

To find the values of \(K_p\) and \(K_c\), we first convert the given amount of SO_3 (5.37 g) to moles (0.0671 mol) and then determine the change in moles at equilibrium. We then calculate the mole fractions and partial pressures at equilibrium, and use these values to find \(K_p\). This involves solving an equation for \(x\), which represents the moles of SO_3 that decompose. Finally, we find \(K_c\) using the relationship with \(K_p\), which in this case is equal to \(K_p\), since the term \(\left(\frac{RT}{P_{\text{atm}}}\right)^{\Delta n}\) becomes 1 due to \(\Delta n = 0\).

Step-by-step solution

Convert grams of \(\mathrm{SO}_{3}\) to moles

We are given 5.37 grams of \(\mathrm{SO}_{3}\). We can use the molar mass of \(\mathrm{SO}_{3}\) to convert this to moles. The molar mass of \(\mathrm{S}\) is about 32 g/mol, and that of \(\mathrm{O}\) is about 16 g/mol. Therefore, the molar mass of \(\mathrm{SO}_{3}\) is \(32 + 3(16) = 80\) g/mol. To convert grams to moles, divide the mass by the molar mass: moles of \(\mathrm{SO}_{3} = \frac{5.37 \text{ g}}{80 \text{ g/mol}} = 0.0671 \text{ mol}\)

Find initial moles of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\)

Initially, there are no moles of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), since the reaction has not yet occurred. Thus, the initial moles of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are both 0.

Determine the change in moles at equilibrium

Let \(x\) be the moles of \(\mathrm{SO}_{3}\) that decompose. Then at equilibrium, there are \(0.0671 - x\) moles of \(\mathrm{SO}_{3}\), \(2x\) moles of \(\mathrm{SO}_{2}\), and \(x\) moles of \(\mathrm{O}_{2}\).

Calculate the mole fractions and partial pressures at equilibrium

Total moles at equilibrium: \(0.0671 - x + 2x + x = 0.0671 + 2x\) Mole fractions at equilibrium: $$ \chi_{\mathrm{SO}_{3}} = \frac{0.0671 - x}{0.0671 + 2x} $$ $$ \chi_{\mathrm{SO}_{2}} = \frac{2x}{0.0671 + 2x} $$ $$ \chi_{\mathrm{O}_{2}} = \frac{x}{0.0671 + 2x} $$ We know that the total pressure is 157 kPa. Thus, we can use the mole fractions to find the partial pressures: $$ P_{\mathrm{SO}_{3}} = \chi_{\mathrm{SO}_{3}} \times 157 \text{ kPa} $$ $$ P_{\mathrm{SO}_{2}} = \chi_{\mathrm{SO}_{2}} \times 157 \text{ kPa} $$ $$ P_{\mathrm{O}_{2}} = \chi_{\mathrm{O}_{2}} \times 157 \text{ kPa} $$

Calculate \(K_p\) using the partial pressures

The expression for \(K_p\) for this reaction is: $$ K_p = \frac{P_{\mathrm{SO}_{2}}^2 \times P_{\mathrm{O}_{2}}}{P_{\mathrm{SO}_{3}}^2} $$ We can substitute the above expressions for the partial pressures in terms of \(x\): $$ K_p = \frac{\left(\frac{2x}{0.0671 + 2x}\times157\right)^2 \times \left(\frac{x}{0.0671 + 2x}\times157\right)}{\left(\frac{0.0671 - x}{0.0671 + 2x}\times157\right)^2} $$

Solve for \(x\)

The expression in step 5 for \(K_p\) in terms of \(x\) can be solved using various methods like trial and error, numerical methods, or algebraically. Solving this equation will give the value of \(x\) which then can be used to find the partial pressures and the equilibrium constant \(K_p\).

Find \(K_c\) using the relationship with \(K_p\)

Once we have the value of \(K_p\), we can find \(K_c\) using the relationship: $$ K_c = K_p \times \left(\frac{RT}{P_{\text{atm}}}\right)^{\Delta n} $$ where \(R = 0.0821 \ \text{L atm/mol K}\) is the ideal gas constant, \(T = 1000 \ \text{K}\), \(\Delta n = 2 - 2 = 0\), and \(P_{\text{atm}} = \frac{157 \ \text{kPa}}{101.325 \ \text{kPa/atm}}\). Since \(\Delta n = 0\), the term \(\left(\frac{RT}{P_{\text{atm}}}\right)^{\Delta n}\) becomes 1. Therefore, \(K_c = K_p\). Once you have the value of \(K_p\) from Step 6, you also have the value of \(K_c\).

Key concepts

Reaction Stoichiometry

Reaction stoichiometry is a fundamental concept in chemistry that involves the quantitative relationship between the reactants and products in a chemical reaction. It tells us how much product can be formed from a given amount of reactant, or vice versa. In our exercise, the balanced chemical equation is \(2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\).

This equation indicates the stoichiometry of the reaction: two moles of sulfur trioxide (\(\mathrm{SO}_{3}\)) decompose to form two moles of sulfur dioxide (\(\mathrm{SO}_{2}\)) and one mole of oxygen gas (\(\mathrm{O}_{2}\)).

Understanding reaction stoichiometry is crucial for determining how the quantities of \(\mathrm{SO}_{3}\), \(\mathrm{SO}_{2}\), and \(\mathrm{O}_{2}\) will change as the reaction proceeds towards equilibrium. This forms the basis for calculating equilibrium concentrations and pressures later in the exercise.

Partial Pressure

Partial pressure is a key concept in the study of gases and equilibrium. It refers to the pressure that a gas in a mixture would exert if it occupied the entire volume by itself at the same temperature. In other words, it is the contribution of each individual gas to the total pressure of the mixture.

In the given problem, at equilibrium, the total pressure in the container is 157 kPa. To find the partial pressures of \(\mathrm{SO}_{3}\), \(\mathrm{SO}_{2}\), and \(\mathrm{O}_{2}\), we need to use the mole fractions of these gases.

Each partial pressure \(P_i\) is calculated by multiplying the total pressure by the mole fraction \(\chi_i\) of that component:

• \(P_{\mathrm{SO}_{3}} = \chi_{\mathrm{SO}_{3}} \times 157\text{ kPa}\)
• \(P_{\mathrm{SO}_{2}} = \chi_{\mathrm{SO}_{2}} \times 157\text{ kPa}\)
• \(P_{\mathrm{O}_{2}} = \chi_{\mathrm{O}_{2}} \times 157\text{ kPa}\)

Partial pressure is essential for calculating the equilibrium constant \(K_p\), which involves the pressures of the reactants and products.

Mole Fraction

Mole fraction is a dimensionless quantity used to express the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a particular substance to the total number of moles present in the mixture.

For the equilibrium calculation, mole fractions are used to determine how much of each gas is present in the mixture relative to the total amount of gas. In our scenario, we compute:

• \(\chi_{\mathrm{SO}_{3}} = \frac{0.0671 - x}{0.0671 + 2x}\)
• \(\chi_{\mathrm{SO}_{2}} = \frac{2x}{0.0671 + 2x}\)
• \(\chi_{\mathrm{O}_{2}} = \frac{x}{0.0671 + 2x}\)

Knowing the mole fractions is essential for calculating partial pressures and understanding the state of equilibrium in the reaction vessel.

Chemical Equilibrium

Chemical equilibrium refers to the state in which the concentrations of reactants and products remain constant over time, as they are being formed and consumed at the same rate.

In the exercise, the decomposition of \(\mathrm{SO}_{3}\) into \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) reaches equilibrium at 157 kPa. The equilibrium constant \(K\) describes the ratio of the concentrations or pressures of products to reactants at this point.

We have two types of equilibrium constants:

• \(K_p\), which uses partial pressures, is expressed as:\[K_p = \frac{P_{\mathrm{SO}_{2}}^2 \times P_{\mathrm{O}_{2}}}{P_{\mathrm{SO}_{3}}^2}\]
• \(K_c\), which is derived from \(K_p\) using the relationship: \(K_c = K_p \times \left(\frac{RT}{P_{\text{atm}}}\right)^{\Delta n}\)

Understanding chemical equilibrium is critical to predicting how the reaction will behave under different conditions and for calculating the values of \(K_p\) and \(K_c\), which remain unchanged as long as the temperature is constant. However, given \(\Delta n = 0\) in our scenario, \(K_c = K_p\).