/*! 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 68 A \(0.831-g\) sample of \(\mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(0.831-g\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00-Lcontainer and heated to \(1100 \mathrm{~K}\). 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 \(1.300 \mathrm{~atm}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Short Answer

Expert verified
Based on the given information and the step-by-step solution, the short answer for determining Kp and Kc is as follows: 1. Calculate initial moles of SO3: \(0.01037 \, \text{mol}\) 2. Set up the ICE table and determine the equilibrium concentrations. 3. Calculate the total pressure and determine x. 4. Use the equilibrium concentrations and partial pressures to write expressions for Kp and Kc: \[K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2}\] \[K_p = \frac{P_{SO_2}^2 P_{O_2}}{P_{SO_3}^2}\] 5. Substitute the equilibrium values into the expressions and calculate the values of Kc and Kp at 1100 K.

Step by step solution

01

Determine initial moles of SO3

To determine the initial moles of SO3, divide the given mass (0.831 g) by its molar mass (80.064 g/mol). \[ \text{moles of SO3} = \frac{0.831 \, \text{g}}{80.064 \, \text{g/mol}} = 0.01037 \, \text{mol} \]
02

Setup the ICE table

Using the initial moles of SO3, we can set up an ICE (Initial, Change, Equilibrium) table to track the changes in the amounts of SO3, SO2 and O2 as the reaction proceeds. The decomposition ratio is 2:2:1. ``` SO3 -> SO2 + 1/2 O2 Initial: 0.01037 mol 0 mol + 0 mol Change: -2x +2x + x Equilibrium: 0.01037-2x 2x x ```
03

Determine equilibrium concentrations

To calculate the equilibrium concentrations and partial pressures, divide the equilibrium amounts of SO3, SO2 and O2 by the container volume (1.00 L). ``` [SO3] = (0.01037-2x) / 1.00\, \text{L} [SO2] = 2x / 1.00\, \text{L} [O2]= x / 1.00\, \text{L} ```
04

Calculate the total pressure and determine x

The total pressure at equilibrium is given as 1.300 atm. That pressure is the sum of the partial pressures of SO3, SO2, and O2. The partial pressures can be calculated using the mole fraction and the ideal gas law. \[P_{total} = P_{SO_3} + P_{SO_2} + P_{O_2}\] Use the ratio of moles and total moles to calculate the individual pressures: \[1.300 \, \text{atm} = \frac{0.01037 - 2x}{0.01037 - x} \times 1.300 \, \text{atm} + \frac{2x}{0.01037 - x} \times 1.300 \, \text{atm} + \frac{x}{0.01037 - x} \times 1.300 \, \text{atm}\] Solve this equation for x and calculate the equilibrium concentrations based on the values of x found.
05

Write expressions for Kp and Kc and calculate their values

Write the expressions for Kp and Kc according to the balanced chemical equation and the equilibrium concentrations and pressures. \[K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2}\] \[K_p = \frac{P_{SO_2}^2 P_{O_2}}{P_{SO_3}^2}\] Now substitute the equilibrium concentrations and partial pressures from Steps 3 and 4 into the above expressions to calculate the values of Kc and Kp for this reaction at 1100 K.

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.

Equilibrium Constant (Kc and Kp)
In chemistry, the quantification of chemical equilibrium is expressed through the equilibrium constant, symbolized as Kc for concentrations and Kp for partial pressures. The equilibrium constant provides insight into the ratio of products to reactants at equilibrium.

For a general reaction, such as \(aA + bB \rightleftharpoons cC + dD\), the concentration-based equilibrium constant (Kc) is defined as \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] where \[\text{[A], [B], [C], [D]}\] represent the molar concentrations of the reactants (A and B) and products (C and D), raised to the power of their respective stoichiometric coefficients (a, b, c, d) in the balanced equation.

Similarly, the pressure-based equilibrium constant (Kp) relates to the partial pressures of gases involved in the reaction and is expressed as \[ K_p = \frac{{(P_{C})}^c{(P_{D})}^d}{{(P_{A})}^a{(P_{B})}^b} \] with \(P_A, P_B, P_C, P_D\) signifying the partial pressures. It's key to note that Kp is only used for reactions involving gases, and its relation to Kc depends on the ideal gas law and the reaction's change in moles of gas.

For the exercise provided, Kp and Kc were calculated based on the equilibrium concentrations and partial pressures obtained from the ICE table and the balanced chemical equation.
ICE Table
An ICE table — standing for Initial, Change, Equilibrium — is an effective method to organize data about the concentrations or pressures of reactants and products in a chemical reaction approaching equilibrium. It simplifies the process of solving for unknown values and is instrumental in equilibrium calculations.

An ICE table begins by listing the initial amounts or concentrations of reactants and products. Then, it shows the change that occurs as the reaction proceeds towards equilibrium, typically represented by the variable 'x'. Finally, the table indicates the equilibrium concentrations or amounts by combining the initial values with the respective changes. In the exercise, through the ICE table, we were able to track the change in the amount of substances from the initial to the equilibrium state and solve for the equilibrium constant.
Partial Pressure
Partial pressure is a crucial concept in understanding gas-phase reactions. It refers to the pressure a single gas component in a mixture would exert if it alone occupied the entire volume. In a mixture of gases, each gas exerts a pressure independent of the others, and the total pressure of the system is the sum of the partial pressures of all gases present.

The importance of partial pressure arises when dealing with gaseous equilibria, as seen in our exercise. According to Dalton's law of partial pressures, the total pressure (P_total) of a mixture of gases is equivalent to the sum of the partial pressures of the individual gases (P_i). Therefore, in the exercise solution, we calculated each gas's partial pressure to solve for the equilibrium constant (Kp), utilizing the ideal gas law.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental guideline predicting how a system at equilibrium responds to external changes. It states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust its equilibrium position to counteract the effect of the change.

For instance, adding more reactants to the system will cause the equilibrium to shift towards products to decrease reactant concentration. Conversely, increasing product concentration prompts the system to produce more reactants. In the context of the given exercise, understanding Le Chatelier's principle can aid in predicting the effect of temperature or pressure changes on the given equilibrium system. However, it's essential to remember that while Le Chatelier's Principle qualitatively predicts the direction of the shift, it does not quantify the amount of change, which is where the equilibrium constants (Kc and Kp) and the ICE table calculations become vital.

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

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 reactionsare \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) 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?

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(\mathrm{~s}) .\) (a) Write the equilibrium- constant expression for this reaction in terms of partial pressures. (b) Why doesn't the concentration of \(\mathrm{Na}_{2} \mathrm{O}\) appear in the equilibrium-constant expression?

If \(K_{c}=1\) for the equilibrium \(2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(\mathrm{g})\), what is the relationship between [A] and [B] at equilibrium?

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate con- stants of \(3.8 \times 10^{-2} \mathrm{~s}^{-1}\) and \(3.1 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B? Explain.

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?

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.