/*! 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 71 Calculate \(\Delta H\) for $$2... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate \(\Delta H\) for $$2 \mathrm{NOCl}(g) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g)$$given the following reactions:$$\begin{array}{ll}\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g) & \Delta H=90.3 \mathrm{~kJ} \\\ \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}(g) & \Delta H=-38.6 \mathrm{~kJ}\end{array}$$

Short Answer

Expert verified
\(\Delta H = 103.4 \) kJ

Step by step solution

01

Identify Given Reactions and Their Enthalpy Changes

The given reactions are: 1) \(\frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}(g) \) with \( \Delta H_1 = 90.3 \text{kJ} \)2) \(\text{NO}(g) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NOCl}(g) \) with \( \Delta H_2 = -38.6 \text{kJ} \)
02

Reverse the Reactions for Formation of \(\text{NOCl} \)

The target reaction needs \( \text{NOCl} \) on the left side. Reverse the second reaction:\(\text{NOCl}(g) \rightarrow \text{NO}(g) + \frac{1}{2} \text{Cl}_2(g)\) with \( \Delta H_2' = +38.6 \text{kJ} \)
03

Combine the Reactions

Add the reactions and their \(\Delta H\) values:\( \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}(g) \) \( \Delta H_1 = 90.3 \text{kJ} \)\( \text{NOCl}(g) \rightarrow \text{NO}(g) + \frac{1}{2} \text{Cl}_2(g) \) \( \Delta H_2' = +38.6 \text{kJ} \)By adding the \(\text{NO}\) terms cancel each other out:
04

Double the Reaction

The target reaction has 2 \( \text{NOCl} \). Multiply the result by 2:\(2 \text{NOCl}(g) \rightarrow \text{N}_2(g) + \text{O}_2(g) + \text{Cl}_2(g) \)and \(\Delta H_{\text{overall}} = 2(51.7) = 103.4 \text{kJ}\)

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.

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.

reaction enthalpy
Reaction enthalpy, denoted as \(\Delta H\), is the heat change associated with a chemical reaction at constant pressure. It indicates whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat). For example, in the given exercise, the enthalpy changes for the reactions are provided. For the first reaction, \(\frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}(g)\), the enthalpy change is +90.3 kJ, indicating an endothermic process. On the other hand, for the reaction \(\text{NO}(g) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NOCl}(g)\), the enthalpy change is -38.6 kJ, indicating an exothermic process.

Understanding these values helps in calculating the overall enthalpy change when combining reactions.
Hess's law
Hess's law states that the total enthalpy change in a chemical reaction is the same regardless of the number of steps the reaction takes. This means you can add up the enthalpy changes of individual steps to find the overall enthalpy change. In the exercise, to find the enthalpy change for \(\text{2 NOCl}(g) \rightarrow \text{N}_2(g) + \text{O}_2(g) + \text{Cl}_2(g)\), we combined two reactions:
  • \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}(g): \Delta H_1 = 90.3 \text{kJ}
  • \text{NO}(g) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NOCl}(g): \Delta H_2 = -38.6 \text{kJ}
By reversing the second reaction and adding the enthalpy changes, we obtained a new reaction: \(\text{NOCl}(g) \rightarrow \text{NO}(g) + \frac{1}{2} \text{Cl}_2(g)\) with an enthalpy change of +38.6 kJ.
thermochemistry
Thermochemistry is the study of the heat involved in chemical processes and reactions. It involves concepts like reaction enthalpy, calorimetry, and Hess's law. For students, it's important to understand how energy is exchanged during reactions.

In the provided exercise, we employed thermochemistry principles to calculate the overall enthalpy change for a complex reaction using given data, showcasing its practical application. By understanding thermochemistry, students can predict how heat transfer influences reaction spontaneity and equilibrium. This field bridges the gap between abstract concepts and real-world chemical reactions, providing insights into both theoretical and practical aspects of chemistry.

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

An unknown volume of water at \(18.2^{\circ} \mathrm{C}\) is added to \(24.4 \mathrm{~mL}\) of water at \(35.0^{\circ} \mathrm{C}\). If the final temperature is \(23.5^{\circ} \mathrm{C},\) what was the unknown volume? (Assume that no heat is released to the surroundings; \(d\) of water \(=1.00 \mathrm{~g} / \mathrm{mL} .\)

Compounds of boron and hydrogen are remarkable for their unusual bonding (described in Section 14.5 ) and for their reactivity. With the more reactive halogens, for example, diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) forms trihalides even at low temperatures: $$\begin{array}{r}\mathrm{B}_{2} \mathrm{H}_{6}(g)+6 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{BCl}_{3}(g)+6 \mathrm{HCl}(g) \\\\\Delta H=-755.4 \mathrm{~kJ}\end{array}$$ What is \(\Delta H\) per kilogram of diborane that reacts?

Draw an enthalpy diagram for a general endothermic reaction; label the axis, reactants, products, and \(\Delta H\) with its sign.

Two iron bolts of equal mass-one at \(100 .{ }^{\circ} \mathrm{C},\) the other at \(55^{\circ} \mathrm{C}\) - are placed in an insulated container. Assuming the heat capacity of the container is negligible, what is the final temperature inside the container \((c\) of iron \(=0.450 \mathrm{~J} / \mathrm{g} \cdot \mathrm{K}) ?\)

Thermal decomposition of 5.0 metric tons of limestone to lime and carbon dioxide absorbs \(9.0 \times 10^{6} \mathrm{~kJ}\) of heat. Convert this energy to (a) joules; (b) calories; (c) British thermal units.

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.