Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 36

Given the following thermochemical equations $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-571.6 \mathrm{~kJ} \\ \mathrm{~N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(l) & & \Delta H=-73.7 \mathrm{~kJ} \\ \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{O}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{HNO}_{3}(l) & & \Delta H=-174.1 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short Answer

Expert verified
The enthalpy change for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at 25°C and 1 atm is -1067.2 kJ.

Step by step solution

01

1. Identify the target reaction

We want to find the enthalpy change for the formation of N₂O₅ from its elements, which means the target reaction is: $$ \frac{1}{2} N_{2}(g) + \frac{5}{2} O_{2}(g) \longrightarrow N_{2}O_{5}(g) $$
02

2. Manipulate the given equations as needed

We'll need to flip the second equation and multiply it by 2, because we want to go from 2 HNO₃ to N₂O₅ + H₂O. For the third equation, we can flip it and multiply by 2 as well. The manipulated equations become: $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-571.6 \mathrm{~kJ} \\ 2 \mathrm{HNO}_{3}(l) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=2(-73.7 \mathrm{~kJ}) \\ \mathrm{HNO}_{3}(l) \longrightarrow \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{O}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g) & & \Delta H=2(-174.1 \mathrm{~kJ}) \end{aligned} $$
03

3. Add the relevant equations

We now add the three manipulated equations: Equation 1: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \qquad\qquad\Delta H=-571.6 \,\text{kJ} $$ Equation 2: $$ 2 \mathrm{HNO}_{3}(l) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l)\qquad\Delta H=2(-73.7) \,\text{kJ} $$ Equation 3: $$ \mathrm{HNO}_{3}(l) \longrightarrow \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{O}_{2}(g)+\frac{1}{2}\mathrm{H}_{2}(g) \qquad\qquad\Delta H=2(-174.1)\,\text{kJ} $$ Sum of the equations: $$ \frac{1}{2} N_{2}(g) + \frac{5}{2} O_{2}(g) \longrightarrow N_{2}O_{5}(g) \qquad\qquad\Delta H = ? $$
04

4. Calculate the enthalpy change for the target reaction

We'll now use Hess's Law to calculate the enthalpy change for the target reaction, by adding the enthalpy changes of the three manipulated equations: $$ \Delta H_{target} = -571.6 + 2(-73.7) + 2(-174.1) = -571.6 -147.4 - 348.2 = -1067.2 \,\text{kJ} $$ So, the enthalpy change for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at \(25 ^{\circ} \mathrm{C}\) and \(1\,\text{atm}\) is \(-1067.2 \,\text{kJ}\).

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

Microwave ovens convert radiation to energy. A microwave oven uses radiation with a wavelength of \(12.5 \mathrm{~cm}\). Assuming that all the energy from the radiation is converted to heat without loss, how many moles of photons are required to raise the temperature of a cup of water \((350.0 \mathrm{~g}\), specific heat \(=4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) ) from \(23.0^{\circ} \mathrm{C}\) to \(99.0^{\circ} \mathrm{C} ?\)

Use the appropriate table to calculate \(\Delta H^{\circ}\) for (a) the reaction between copper(II) oxide and carbon monoxide to give copper metal and carbon dioxide. (b) the decomposition of one mole of methyl alcohol (CH \(_{3} \mathrm{OH}\) ) to methane and oxygen gases.

When one mole of caffeine \(\left(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\right)\) is burned in air, \(4.96 \times 10^{3} \mathrm{~kJ}\) of heat is evolved. Five grams of caffeine is burned in a bomb calorimeter. The temperature is observed to increase by \(11.37^{\circ} \mathrm{C}\). What is the heat capacity of the calorimeter in \(\mathrm{J} /{ }^{\circ} \mathrm{C} ?\)

Draw a cylinder with a movable piston containing six molecules of a liquid. A pressure of 1 atm is exerted on the piston. Next draw the same cylinder after the liquid has been vaporized. A pressure of one atmosphere is still exerted on the piston. Is work done on the system or by the system?

A lead ore, galena, consisting mainly of lead(II) sulfide, is the principal source of lead. To obtain the lead, the ore is first heated in the air to form lead oxide. $$ \mathrm{PbS}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{PbO}(s)+\mathrm{SO}_{2}(g) \quad \Delta H=-415.4 \mathrm{~kJ} $$ The oxide is then reduced to metal with carbon. $$ \mathrm{PbO}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Pb}(s)+\mathrm{CO}(g) \quad \Delta H=+108.5 \mathrm{k}] $$ Calculate \(\Delta H\) for the reaction of one mole of lead(II) sulfide with oxygen and carbon, forming lead, sulfur dioxide, and carbon monoxide.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

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