Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 5.8, Problem 1PPA
Interpretation Introduction
Interpretation:
The standard enthalpy change of the reaction is to be calculated.
Concept introduction:
The enthalpy of a reactionthattakes place at standard conditions is known as the standard enthalpy of the reaction.
The standard enthalpy of formation of a compound is defined as the amount of heat change involved when 1mole of a particular compound is formed from its constituent elements in their standard states.
The standard enthalpy of formation of an element in its standard or most stable state is zero.
Enthalpy change of a reaction is the difference of “sum of enthalpy change of the products” and “sum of enthalpy change of the reactants.”
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A 115.26 g sample of aluminum is heated from 37 oC to its melting point of 660 oC and then melted completely. How much energy was absorbed by the sample? (Answer in kJ)
S=0.91 J/goC Delta Hf = 10.7 kJ/mol
QUESTION 8
Determine ΔH for the third reaction from the information given.
2 CO2(g)
↔ 2 CO(g) + O2(g)
ΔH = 566.0 kJ
3 CO(g) + O3(g)
↔ 3 CO2
ΔH = -992.0 kJ
3/2 O2(g)
↔ O3(g)
ΔH = ?
Key Concept: Hess's law. The enthalpy change of an overall reaction is simply the sum of the enthalpy changes for individual steps that give the overall reaction.
Practice Exercise 1If the heat of formation of H2O1l2 is -286 kJ>mol, whichof the following thermochemical equations is correct?(a) 2 H1g2 + O1g2¡H2O1l2 ΔH = -286 kJ(b) 2 H21g2 + O21g2¡2 H2O1l2 ΔH = -286 kJ(c) H21g2 + 12 O21g2¡H2O1l2 ΔH = -286 kJ(d) H21g2 + O1g2¡H2O1g2 ΔH = -286 kJ(e) H2O1l2¡H21g2 + 12O21g2 ΔH = -286 kJ
Chapter 5 Solutions
Chemistry
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