/*! 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 70 The vapor pressure of water at \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(0.0313\) atm. Calculate the values of \(K_{\mathrm{p}}\) and \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
\(K_p = 0.0313\) atm; \(K_c \approx 0.00128\) mol/L.

Step by step solution

01

Understanding the Reaction

The equilibrium for the reaction is given as: \( \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \). In this process, liquid water is turning into its gaseous form. Since water vapor is a gas, its contribution to the equilibrium expression is set using the given vapor pressure.
02

Calculating Kp (Equilibrium Constant in Terms of Pressure)

For the equilibrium \( \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \), the equilibrium constant \( K_p \) is the partial pressure of water vapor at equilibrium. Given that the vapor pressure of water at this temperature is \( 0.0313 \) atm, \( K_p \) can be calculated directly: \( K_p = 0.0313 \) atm.
03

Using the Ideal Gas Law for Kc

The equilibrium constant \( K_c \) is related to \( K_p \) by the expression \( K_p = K_c (RT)^{\Delta n} \), where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas, which is 1 in this case (since liquid to gas transition provides a mole change from 0 to 1).
04

Converting Temperature to Kelvin

Convert the temperature from degrees Celsius to Kelvin: \( T = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \) K.
05

Solving for Kc

Using the relation \( K_p = K_c (RT) \), rearrange to find \( K_c \): \( K_c = \frac{K_p}{RT} \). Use \( R = 0.0821 \) L atm / mol K: \( K_c = \frac{0.0313}{(0.0821)(298.15)} \approx 0.00128 \) mol/L.

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.

Vapor Pressure
Vapor pressure is an important concept when discussing phase equilibrium and changes of state.
In simple terms, vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid form in a closed system. This occurs when evaporation and condensation rates are equal.

When you hear about a liquid's vapor pressure, it's telling you how much of that liquid can turn into gas at a given temperature. Water, for instance, has a vapor pressure of 0.0313 atm at 25°C. That means at this temperature, the gas molecules of water have a pressure of 0.0313 atm above the liquid.

Vapor pressure increases with temperature since more molecules have sufficient energy to escape into the vapor phase. Understanding this is crucial for calculating equilibrium constants such as \(K_p\), which depend on the vapor pressure of substances involved.
Ideal Gas Law
The ideal gas law is a vital tool in chemistry for linking pressure, volume, and temperature with the amount of gas present. The formula is expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is the universal gas constant, and \(T\) is temperature in Kelvin.

This law assumes gases behave ideally, meaning that they follow this equation at all conditions. However, in reality, gases may deviate at high pressures or low temperatures. For typical problems, we use \(R = 0.0821\) L atm / mol K.

In the equilibrium expression, the ideal gas law helps connect \(K_p\) and \(K_c\), where \(K_p\) is related to pressure and \(K_c\) relates to concentration (mol/L):
  • \(K_p = K_c(RT)^{\Delta n}\)
  • \({\Delta n}\) is the change in moles of gas, important for reactions involving phase changes.
Temperature Conversion
Temperature conversion is crucial in problem-solving, especially for equilibrium calculations where temperature plays a key role.
The most common conversion you need is Celsius to Kelvin: \\[ T(K) = T(°C) + 273.15 \]

Kelvin is the standard unit for scientific calculations because it starts at absolute zero, making it directly proportional to the kinetic energy of particles. This ensures consistent results when using the ideal gas law and calculating equilibrium constants.

In the given exercise, converting 25°C to Kelvin (298.15 K) allows us to correctly calculate \(K_c\) using the ideal gas law relation to \(K_p\).
Phase Equilibrium
Phase equilibrium is the condition where different phases of a substance coexist stably.
For example, in the case of water, both the liquid and gas phases are present at equilibrium, with no net change between them.

At equilibrium, processes like evaporation and condensation occur at equal rates. This balance determines the vapor pressure and influences equilibrium constants. In our exercise, the phase equilibrium between liquid and gaseous water leads to a vapor pressure of 0.0313 atm.

Understanding phase equilibrium helps in predicting how a substance will behave under different conditions, such as temperature changes. It is also critical for calculating \(K_p\) and \(K_c\), which represent the extent of a reaction at phase equilibrium.

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

Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation $$ 2 \mathrm{ClO}(g) \rightleftharpoons \mathrm{Cl}_{2} \mathrm{O}_{2}(g) $$ for which \(K_{c}\) is \(4.96 \times 10^{11}\) at \(253 \mathrm{~K}\). For an equilibrium mixture in which \(\left[\mathrm{Cl}_{2} \mathrm{O}_{2}\right]\) is \(6.00 \times 10^{-6} \mathrm{M}\), what is \([\mathrm{ClO}] ?\)

Consider the reaction \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) .\) When \(1.50 \mathrm{~mol}\) of \(\mathrm{CO}_{2}\) and an excess of solid carbon are heated in a \(20.0 \mathrm{~L}\) container at \(1100 \mathrm{~K}\), the equilibrium concentration of \(\mathrm{CO}\) is \(7.00 \times 10^{-2} \mathrm{M}\) (a) What is the equilibrium concentration of \(\mathrm{CO}_{2}\) ? (b) What is the value of the equilibrium constant \(K_{c}\) at \(1100 \mathrm{~K} ?\)

For the reaction \(2 \mathrm{~A}_{3}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{~A}_{3} \mathrm{~B}\), the rate of the for- ward reaction is \(0.35 \mathrm{M} / \mathrm{s}\) and the rate of the reverse reaction is \(0.65 \mathrm{M} / \mathrm{s}\). The reaction is not at equilibrium. Will the reaction proceed in the forward or reverse direction to attain equilibrium?

Solid particles that form in the kidney are called kidney stones and frequently cause acute pain. One common type of kidney stone is formed from a precipitation reaction of calcium and oxalate: $$ \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{CaC}_{2} \mathrm{O}_{4}(s) $$ Use Le Châtelier's principle to explain the following statements. (a) A person taking diuretics, medicines that help kidneys remove fluids, may be at increased risk for developing kidney stones. (b) A person diagnosed with hypercalciuria, a genetic condition causing elevated levels of calcium in the urine, has an increased risk for developing kidney stones. (c) One simple treatment for kidney stones is to avoid foods high in oxalate such as spinach, rhubarb, and nuts. (d) Another simple treatment for kidney stones is to increase consumption of water.

Equilibrium constants at \(25^{\circ} \mathrm{C}\) are given for the sequential equilibrium reactions in the binding of oxygen to hemoglobin. \(\mathrm{Hb}+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right) \quad \mathrm{K}_{\mathrm{cl}}=1.5 \times 10^{4}\) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{2} \quad \mathrm{~K}_{\mathrm{c} 2}=3.5 \times 10^{4}\) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{3} \quad \mathrm{~K}_{\mathrm{c} 3}=5.9 \times 10^{4}\) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{3}+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4} \quad \mathrm{~K}_{\mathrm{c} 4}=1.7 \times 10^{6}\) (a) How does the value of the equilibrium constant change for each sequential binding step \(\left(K_{c 1}\right.\) through \(K_{c 4}\) )? (b) Do the equilibrium reactions become more product-or reactant-favored with the binding of each oxygen? (c) How do the sequential equilibrium steps affect the oxygencarrying capacity of hemoglobin?

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.