The equilibrium constant for the formation of NiO at 1627 ° C should be calcualted under given conditions. Concept introduction: The Gibbs free energy or the free energy change is a thermodynamic quantity represented by Δ r G o . It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is: Δ r G ° = ∑ nΔ f G ° ( products ) - ∑ nΔ f G ° ( reactants ) Δ f G o is related to the equilibrium constant K by the equation, Δ f G o = -RTlnK p The rearranged expression is, K p = e - Δ f G o RT Δ f G o is also related to the reaction quotient Q by the expression, Δ f G = Δ f G ° +RTlnQ For a general reaction, aA + bB → cC + dD Q = [ C ] c [ D ] d [ A ] a [ B ] b
The equilibrium constant for the formation of NiO at 1627 ° C should be calcualted under given conditions. Concept introduction: The Gibbs free energy or the free energy change is a thermodynamic quantity represented by Δ r G o . It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is: Δ r G ° = ∑ nΔ f G ° ( products ) - ∑ nΔ f G ° ( reactants ) Δ f G o is related to the equilibrium constant K by the equation, Δ f G o = -RTlnK p The rearranged expression is, K p = e - Δ f G o RT Δ f G o is also related to the reaction quotient Q by the expression, Δ f G = Δ f G ° +RTlnQ For a general reaction, aA + bB → cC + dD Q = [ C ] c [ D ] d [ A ] a [ B ] b
Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.
Chapter 18, Problem 82IL
Interpretation Introduction
Interpretation:
The equilibrium constant for the formation of NiO at 1627 °C should be calcualted under given conditions.
Concept introduction:
The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:
ΔrG°= ∑nΔfG°(products)- ∑nΔfG°(reactants)
ΔfGo is related to the equilibrium constant K by the equation,
ΔfGo= -RTlnKp
The rearranged expression is,
Kp= e-ΔfGoRT
ΔfGo is also related to the reaction quotient Q by the expression,
1. Determine the relationship between the following molecules as identical, diastereomers, or enantiomers (6
points, 2 points each).
OH
OH
OH
A-A
OH
HOT
HO-
ACHN
and
HO-
ACHN
OH
HO
HO
°
OH
and
OH
OH
SH
and
...SH
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please provide the structure for this problem, thank you!
Chapter 18 Solutions
Owlv2 With Ebook, 1 Term (6 Months) Printed Access Card For Kotz/treichel/townsend/treichel's Chemistry & Chemical Reactivity, 10th
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The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY