Ch. 17 - Chemical ThermodynamicsWorksheetSee all chapters
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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: If ΔH = -80.0 kJ and ΔS = -0.400 kJ/K, the reaction is spontaneous below a certain temperature. Calculate that temperature. Express your answer numerically in kelvins.

Solution: If ΔH = -80.0 kJ and ΔS = -0.400 kJ/K, the reaction is spontaneous below a certain temperature. Calculate that temperature. Express your answer numerically in kelvins.

Problem

If ΔH = -80.0 kJ and ΔS = -0.400 kJ/K, the reaction is spontaneous below a certain temperature. Calculate that temperature. Express your answer numerically in kelvins.


Solution

We’re being asked to determine at what temperature the given reaction is spontaneous. The standard free energy change of a reaction (ΔG) is given by the following equation:



Recall that if:

• ΔG < 0 or ΔG = (–); the reaction is spontaneous

• ΔG = 0; the reaction is at equilibrium

• ΔG > 0 or ΔG = (+); the reaction is non-spontaneous


We first need to determine the temperature at which the reaction is at equilibrium. We’re given ΔH = –80.0 kJ and ΔS = –0.400 kJ/K for the reaction. This means:



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