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Chapter 15: Problem 46

At \(850 \mathrm{~K},\) the following reaction has \(K_{p}=0.0035:\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ In an equilibrium mixture the partial pressures of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are \(18.24 \mathrm{kPa}\) and \(50.66 \mathrm{kPa}\), respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

Short Answer

Expert verified
The equilibrium partial pressure of \( \text{SO}_3 \) is approximately 7.686 kPa.

Step by step solution

01

Write the Expression for Kp

The equilibrium constant, \( K_p \), for the reaction \( 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \) is given by the expression: \[ K_p = \frac{{P_{\text{SO}_3}^2}}{{P_{\text{SO}_2}^2 \cdot P_{\text{O}_2}}} \] Here, \( P_{\text{SO}_3} \), \( P_{\text{SO}_2} \), and \( P_{\text{O}_2} \) are the equilibrium partial pressures of \( \text{SO}_3 \), \( \text{SO}_2 \), and \( \text{O}_2 \) respectively. The given \( K_p \) is 0.0035.
02

Substitute Known Values

We know that \( P_{\text{SO}_2} = 18.24 \text{kPa} \), \( P_{\text{O}_2} = 50.66 \text{kPa} \), and \( K_p = 0.0035 \). Substitute these values into the expression for \( K_p \): \[ 0.0035 = \frac{{P_{\text{SO}_3}^2}}{{(18.24)^2 \cdot 50.66}} \]
03

Solve for P_SO3

To find \( P_{\text{SO}_3} \), rearrange the equation to solve for it:\[ P_{\text{SO}_3}^2 = 0.0035 \times (18.24)^2 \times 50.66 \] Calculate the right-hand side:\[ P_{\text{SO}_3}^2 = 0.0035 \times 332.6976 \times 50.66 \] \[ P_{\text{SO}_3}^2 = 59.0511 \] Now take the square root to find \( P_{\text{SO}_3} \):\[ P_{\text{SO}_3} = \sqrt{59.0511} \approx 7.686 \text{kPa} \]

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as \( K_p \) for reactions involving gases, represents the ratio of product pressures to reactant pressures when a reaction is at equilibrium. When dealing with chemical equilibrium in gases, \( K_p \) is calculated using the partial pressures of the gases involved.For the given chemical reaction \( 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \):
  • \( K_p = \frac{{P_{\text{SO}_3}^2}}{{P_{\text{SO}_2}^2 \cdot P_{\text{O}_2}}} \)
This equation shows that the products' pressures are divided by the reactants' pressures, adjusted for their coefficients in the balanced chemical equation. Each pressure term is raised to a power equal to its respective coefficient in the balanced equation.In our problem, \( K_p = 0.0035 \) at a temperature of \( 850 \mathrm{~K} \). This value indicates the degree to which the reactants are converted into products at equilibrium.A small \( K_p \) value, like in our scenario, suggests that, at equilibrium, the reactants are more prevalent than the products.
Partial Pressure
Partial pressure refers to the pressure that a single gas in a mixture contributes to the total pressure.Each gas in a mixture behaves as if it occupies the entire volume independently, contributing its own pressure, without affecting the other gases. For example, in the reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\),the partial pressures of \( \text{SO}_2 \) and \( \text{O}_2 \) were given as \( 18.24 \text{kPa} \) and \( 50.66 \text{kPa} \) respectively.Knowing the partial pressures of all other components in the reaction allows us to solve for the unknown \( \text{SO}_3 \) partial pressure using the given \( K_p \) equation.
Le Ch芒telier's Principle
Le Ch芒telier's Principle helps us predict how changes in conditions can shift the equilibrium position of a chemical reaction.This principle states that if a dynamic equilibrium is disturbed by changing the conditions,
  • The system will adjust itself to partially counteract the effect of the change and a new equilibrium will be established.
Consider changes such as temperature, pressure, or concentration when thinking about the gas reaction:\(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\).If the pressure is increased, the system will shift to reduce this change by favoring the side of the reaction with fewer gas molecules.In our equation, increasing pressure would favor the production of \( \text{SO}_3 \), which has fewer moles of gas on the product side compared to its reactants.Thus, understanding Le Ch芒telier's Principle helps explain how external changes might affect the equilibrium conditions and outcomes.

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Most popular questions from this chapter

How do the following changes affect the value of the \(K_{P}\) for a gas-phase endothermic reaction: \((\mathbf{a})\) increase in the total pressure by adding a noble gas,(b) addition of a reactant, (c) increase in the temperature (d) increase in the volume, \((\mathbf{e})\) decrease in the temperature?

Suppose that the gas-phase reactions \(A \longrightarrow B\) and \(B \longrightarrow A\) are both elementary processes with rate con- stants of \(2.5 \times 10^{-2} \mathrm{~min}^{-1}\) and \(2.5 \times 10^{-1} \mathrm{~min}^{-1}\), respectively. (a) What is the value of the equilibrium constant (b) Which is greater for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) ? at equilibrium, the partial pressure of A or the partial pressure of B?

At a temperature of \(700 \mathrm{~K},\) the forward and reverse rate constants for the reaction \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\) are \(k_{f}=1.8 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(k_{r}=0.063 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (a) What is the value of the equilibrium constant \(K_{c}\) at \(700 \mathrm{~K} ?(\mathbf{b})\) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of \(k_{f}=0.097 M^{-1} \mathrm{~s}^{-1}\) and \(k_{r}=2.6 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(800 \mathrm{~K}\) ?

For the reaction, at \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 10.0-L flask containing an equilibrium mixture of the three gases, there are \(1.30 \mathrm{~g} \mathrm{H}_{2}\) and \(21.0 \mathrm{~g} \mathrm{I}_{2}\). What is the mass of HI in the flask?

The following equilibria were measured at \(823 \mathrm{~K}\) : $$ \begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array} $$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c}\), for the reaction \(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K} .(\mathbf{b})\) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of \(101.3 \mathrm{kPa}\) and a temperature of \(298 \mathrm{~K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). \((\mathbf{d})\) If the reaction vessel from part (c) is heated to \(823 \mathrm{~K}\) and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?
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