Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 100

In Section \(11.5,\) we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for $K_{p \cdot}(\mathbf{b})\( By using data in Appendix \)\mathrm{B}$, give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C}\). (c) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

Short Answer

Expert verified
The equilibrium equation between liquid water and water vapor is: \( H_2O(l) \rightleftharpoons H_2O(g) \). The expression for the equilibrium constant, Kp, is: \( K_p = P_{H_2O(g)} \). At 30°C, the vapor pressure of water is 4.24 kPa, so Kp is 4.24 kPa. At the normal boiling point of any liquid, the value of Kp is equal to the atmospheric pressure, which is 101.3 kPa.

Step by step solution

01

Write the equilibrium equation for liquid water and water vapor

We start by writing the equilibrium equation between liquid water and water vapor. The reaction can be represented as: \( H_2O(l) \rightleftharpoons H_2O(g) \) Here, \(H_2O(l)\) represents liquid water and \(H_2O(g)\) represents water vapor.
02

Write the expression for Kp

Since we have the chemical reaction for the equilibrium process, we can now write the expression for the equilibrium constant, Kp. For this reaction, we have: \( K_p = \frac{[H_2O(g)]}{[H_2O(l)]} \) However, as Kp only depends on gaseous species, we need to consider partial pressures of the gases in the expression. Since the liquid water is not in the gaseous phase, its partial pressure has no effect on Kp. Thus, the expression for Kp becomes: \( K_p = P_{H_2O(g)} \)
03

Find the value of Kp at 30°C

Now that we have the expression for Kp, we can find its value at 30°C using the data in Appendix B. In the appendix, we find the vapor pressure of water at various temperatures. At 30°C, the vapor pressure of water is given as 4.24 kPa. So, the value of Kp is: \( K_p = P_{H_2O(g)} = 4.24 \ kPa \)
04

Find the value of Kp at the normal boiling point

We are asked to find the value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid. At normal boiling point, the vapor pressure of the liquid is equal to the atmospheric pressure. For this exercise, we assume that atmospheric pressure is 1 atm (101.3 kPa). Therefore, at the normal boiling point, the value of Kp is: \( K_p = P= 101.3 \ kPa \) So, the value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid is 101.3 kPa.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (a) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

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}\( for this reaction. (a) If excess \)\operatorname{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 \(1.4 \mathrm{~L},\) what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

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 \(1100 \mathrm{~K}\).

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)

The equilibrium $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\( is established at \)550 \mathrm{~K}$. An equilibrium mixture of the three gases has partial pressures of $10.13 \mathrm{kPa}, 20.27 \mathrm{kPa}\(, and \)35.46 \mathrm{kPa}\( for \)\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.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation
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.

Sign-up for free