Ch.6 - Thermochemistry WorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Reaction of gaseous ClF with F 2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH°rxn for this reaction:(1) 2CIF(g) + O2(g) ⟶ Cl2O(g) + O

Solution: Reaction of gaseous ClF with F 2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH°rxn for this reaction:(1) 2CIF(g) + O2(g) ⟶ Cl2O(g) + O

Problem

Reaction of gaseous ClF with F 2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH°rxn for this reaction:

(1) 2CIF(g) + O2(g) ⟶ Cl2O(g) + OF2(g)          ΔH°rxn = 167.5 kJ
(2) 2F2(g) + O2(g) ⟶ 2OF2(g)                         ΔH°rxn = −43.5 kJ
(3) 2ClF3(l) + 2O2(g) ⟶ Cl2O(g) + 3OF2(g)    ΔH°rxn = 394.1 kJ

2ClF  + 2F2 ⟶  2ClF3

Solution

To find the ΔHrxn, we need to use the Hess's Law where we have to rearrange the given equations to get the desired reaction.

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