At \(900 \mathrm{~K}\) the following reaction has \(K_{p}=0.345\) : $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

Short Answer

Expert verified
The equilibrium partial pressure of \(SO_{3}\) in the mixture is approximately 0.05018 atm.

Step by step solution

01

Write the equilibrium constant expression

The equilibrium constant expression for the reaction is: \[K_{p} = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2} \cdot [O_{2}]}\]
02

Substitute the given values

We are given that \(K_p = 0.345\), \([SO_2] = 0.135 \mathrm{~atm}\), and \([O_2] = 0.455 \mathrm{~atm}\). Let's denote the equilibrium partial pressure of \(SO_3\) as P. With these values, our equation becomes: \[0.345 = \frac{P^{2}}{(0.135)^{2} \cdot 0.455}\]
03

Solve for the equilibrium partial pressure of SO₃

To find the equilibrium partial pressure of SO₃ (P), we will first isolate P² by multiplying both sides of the equation by the denominator (\((0.135)^{2} \cdot 0.455)\)): \[P^{2} = 0.345 \times (0.135)^{2} \cdot 0.455\] Now, calculate the value of P²: \[P^{2} \approx 0.002518\] Finally, take the square root of both sides to find the equilibrium partial pressure of SO₃: \[P = \sqrt{0.002518} \approx 0.05018 \mathrm{~atm}\]
04

Conclusion

The equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture is approximately 0.05018 atm.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial in the study of reactions and their behavior under different conditions. Chemical equilibrium refers to the state of a reaction wherein the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the concentration of reactants and products over time. It's a dynamic condition, meaning that the molecules are constantly reacting, but the overall concentrations remain constant.

At equilibrium, the system has reached a state of balance, but it doesn't mean that the reactants and products are at equal concentrations. Instead, they are at a ratio that will remain consistent as long as external conditions such as temperature or pressure do not change. External changes can 'shift' the equilibrium position according to Le Chatelier's Principle, which predicts how the position of equilibrium will change to counteract the effect of the disturbance.
Equilibrium Constant Expression
The equilibrium constant expression for a reaction is a way of quantifying the position of equilibrium. It is symbolized by the letter 'K'. For reactions involving gases, we use partial pressures and represent the constant as \(K_p\). The equilibrium constant is a reflection of the proportions of the concentrations of the products to the reactants at equilibrium and each is raised to the power of their coefficients in the balanced chemical equation.

To construct an equilibrium constant expression, the concentrations of gaseous products are divided by the concentrations of gaseous reactants, taking into account their stoichiometric coefficients from the balanced equation. The value of \(K\) is constant for a given reaction at a specific temperature. When the system is at equilibrium, the constant expression will yield the equilibrium constant's value, which can provide insights into the extent of the reaction and the proportions of reactants and products.
Partial Pressure Calculation
Partial pressure is a measure of the individual pressure exerted by a specific gas in a mixture of gases. It is proportional to the mole fraction of the gas in the total mixture. To calculate partial pressures, we make use of Dalton's Law, which states that the total pressure of a mixture of gases is the sum of the partial pressures of each individual gas.

In an equilibrium situation involving gases, the equilibrium constant expression \(K_p\) is particularly useful. By rearranging the equilibrium expression and substituting the known values of partial pressures for the reactants, as well as the value of \(K_p\), it is possible to calculate the partial pressure of an unknown product. This involves algebraic manipulation to first isolate the term that includes the unknown partial pressure and then solving for that unknown, often by taking square roots if the term is squared, as in the case of our given exercise.

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Most popular questions from this chapter

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned}\mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v)\rightleftharpoons & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(s o l v)+\mathrm{H}_{2} \mathrm{O}(so l v)\end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is \(6.68 .\) A pharmaceutical chemist makes up \(15.0 \mathrm{~L}\) of a solution that is initially \(0.275 \mathrm{M}\) in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is estab- lished at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\), respectively. (a) Calculate \(K_{p}\) for this reaction at \(500.0 \mathrm{~K}\). (b) If the vessel has a volume of \(5.00 \mathrm{~L}\) calculate \(K_{c}\) at this temperature.

Water molecules in the atmosphere can form hydrogenbonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSEPR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at \(300 \mathrm{~K}\) and 0.020 at \(350 \mathrm{~K}\). Is water dimer formation endothermic or exothermic?

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C}\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium? (c) What additional information would you need in order to decide whether the reaction as written is endothermic or exothermic?

As shown in Table \(15.2, K_{p}\) for the equilibrium $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.51 \times 10^{-5}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C}\). If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) \(98 \mathrm{~atm} \mathrm{NH}_{3}, 45 \mathrm{~atm} \mathrm{~N}_{2}, 55 \mathrm{~atm} \mathrm{H}_{2}\) (b) \(57 \mathrm{~atm} \mathrm{NH}_{3}, 143 \mathrm{~atm} \mathrm{~N}_{2},\) no \(\mathrm{H}_{2}\) (c) \(13 \mathrm{~atm} \mathrm{NH}_{2} 27 \mathrm{~atm} \mathrm{~N}_{2} 82 \mathrm{~atm} \mathrm{H}_{2}\)

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