/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 At \(900 \mathrm{~K}\) the follo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(0.135 \mathrm{~atm}\) and \(0.455 \mathrm{~atm}\), respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

Short Answer

Expert verified
The equilibrium partial pressure of SO3 in the mixture is approximately \(0.3276\) atm.

Step by step solution

01

Write the Kp expression for the given reaction

For the reaction: \(2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)\), The equilibrium constant expression (Kp) can be written as: \(K_p = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\), where [SO3], [SO2], and [O2] represent the equilibrium partial pressures of SO3, SO2, and O2, respectively.
02

Substitute the given values

We are given the following values: \( K_p = 0.345 \) \[SO_2 = 0.135 atm\] \[O_2 = 0.455 atm\] We will substitute these values into the Kp expression: \(0.345 = \frac{[SO_3]^2}{(0.135)^2(0.455)}\)
03

Solve for the equilibrium partial pressure of SO3

Now, we need to solve this equation for [SO3]: \([SO_3]^2 = (0.345)((0.135)^2(0.455))\) \([SO_3] = \sqrt{(0.345)((0.135)^2(0.455))}\) \([SO_3] = 0.3276 atm\) The equilibrium partial pressure of SO3 in the mixture is approximately \(0.3276\) atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a reversible chemical reaction where the rates of the forward reaction and the reverse reaction are equal. This balance between the two reactions results in constant concentrations or partial pressures of the reactants and products when the system is closed and at a constant temperature.

For example, in the reaction provided, sulfur dioxide (SO2) and oxygen (O2) react to form sulfur trioxide (SO3). At equilibrium, the formation of SO3 from SO2 and O2 occurs at the same rate as the decomposition of SO3 back into SO2 and O2. This dynamic state can be disturbed by changes in concentration, pressure, volume, or temperature, resulting in a shift in the equilibrium position according to Le Chatelier's principle.
Equilibrium Constant (Kp)
The equilibrium constant for partial pressures, denoted as Kp, is a dimensionless number that provides a quantitative measure of the position of equilibrium for a gaseous system. It is determined by the ratio of the products' partial pressures to the reactants' partial pressures, each raised to the power of their stoichiometric coefficients.

In the given reaction, the Kp expression is \(K_p = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\). Each partial pressure is represented by brackets and raised to the power corresponding to its coefficient in the balanced chemical equation. For this reaction at 900 K, Kp is given to be 0.345. This value is key to determining the equilibrium partial pressures of the involved gases.
Partial Pressure
Partial pressure refers to the pressure that a single gas component in a mixture of gases would exert if it occupied the entire volume alone at the same temperature. Dalton’s Law of Partial Pressures states that the total pressure of a mixture of gases is the sum of the partial pressures of the individual gases.

In the context of the provided exercise, when the reaction reaches equilibrium at 900 K, the partial pressures of SO2 and O2 are given, and we use these values along with the Kp to calculate the partial pressure of SO3. Understanding how to calculate and interpret partial pressures is essential in solving equilibrium problems involving gaseous systems. It's also critical in applications such as determining the composition of atmospheric gases or the conditions for industrial chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\)

At \(100^{\circ} \mathrm{C}\) the equilibrium constant for the reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) has the value \(K_{c}=\) \(2.19 \times 10^{-10}\). Are the following mixtures of \(\mathrm{COCl}_{2}, \mathrm{CO}\), and \(\mathrm{Cl}_{2}\) at \(100^{\circ} \mathrm{C}\) at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) \(\left[\mathrm{COCl}_{2}\right]=2.00 \times 10^{-3} \mathrm{M}, \quad[\mathrm{CO}]=3.3 \times 10^{-6} \mathrm{M}\) \(\left[\mathrm{Cl}_{2}\right]=6.62 \times 10^{-6} M ; \quad\) (b) \(\left[\mathrm{COCl}_{2}\right]=4.50 \times 10^{-2} M\) \([\mathrm{CO}]=1.1 \times 10^{-7} \mathrm{M},\left[\mathrm{Cl}_{2}\right]=2.25 \times 10^{-6} \mathrm{M} ;\) (c) \(\left[\mathrm{COCl}_{2}\right]=\) \(0.0100 M,[\mathrm{CO}]=\left[\mathrm{Cl}_{2}\right]=1.48 \times 10^{-6} \mathrm{M}\)

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(\mathrm{~g})\) ?

Write the equilibrium-constant expression for the equilibrium $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ The table included below shows the relative mole percentages of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{CO}(\mathrm{g})\) at a total pressure of \(1 \mathrm{~atm}\) for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic? Explain. $$ \begin{array}{lll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \mathrm{CO}_{2}(\text { mol } \%) & \text { CO (mol \%) } \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\ 1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\ \hline \end{array} $$

Consider the reaction $$ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) $$ At \(25^{\circ} \mathrm{C}\) the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess \(\mathrm{CaSO}_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4}\), what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(3.0 \mathrm{~L}\), what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.