Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 15, Problem 113E
Interpretation Introduction
Interpretation:
The reason for the formation of dew overnight is to be stated.
Concept introduction:
The partial pressure exerted by the particles present in the open space above any liquid is termed as vapor pressure. When the gas space above the liquid is closed, the vapor pressure increases to a value known as the equilibrium vapor pressure.
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MISSED THIS? Read Section 19.9 (Pages 878-881); Watch IWE 19.10
Consider the following reaction:
CH3OH(g)
CO(g) + 2H2(g)
(Note that AG,CH3OH(g) = -162.3 kJ/mol and AG,co(g)=-137.2 kJ/mol.)
Part A
Calculate AG for this reaction at 25 °C under the following conditions:
PCH₂OH
Pco
PH2
0.815 atm
=
0.140 atm
0.170 atm
Express your answer in kilojoules to three significant figures.
Ο ΑΣΦ
AG = -150
Submit
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kJ
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Calculate the free energy change under nonstandard conditions (AGrxn) by using the following relationship:
AGrxn = AGrxn + RTInQ,
AGxn+RTInQ,
where AGxn is the standard free energy change, R is the ideal gas constant, T is the temperature in kelvins, a
is the reaction quotient.
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Chapter 15 Solutions
Introductory Chemistry: An Active Learning Approach
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