The free energy Δ G o of the given compound is to be calculated using the K s p value of P b C l 2 . Concept introduction: Solubility product: The products of power of ion concentrations, which are dissolved in solvent is known as solubility product and it is it is also known as equilibrium between solute (ionic solids) and its saturated solutions. Free energy change ΔG : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant K ΔG = ΔG o + RT ln ( K ) ΔG o = ΔH o − TΔS o Where, T is the temperature ΔG is the free energy ΔG o , ΔH o and ΔS o is standard free energy, enthalpy and entropy values.
The free energy Δ G o of the given compound is to be calculated using the K s p value of P b C l 2 . Concept introduction: Solubility product: The products of power of ion concentrations, which are dissolved in solvent is known as solubility product and it is it is also known as equilibrium between solute (ionic solids) and its saturated solutions. Free energy change ΔG : change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant K ΔG = ΔG o + RT ln ( K ) ΔG o = ΔH o − TΔS o Where, T is the temperature ΔG is the free energy ΔG o , ΔH o and ΔS o is standard free energy, enthalpy and entropy values.
The free energy ΔGo of the given compound is to be calculated using the Ksp value of PbCl2.
Concept introduction:
Solubility product: The products of power of ion concentrations, which are dissolved in solvent is known as solubility product and it is it is also known as equilibrium between solute (ionic solids) and its saturated solutions.
Free energy changeΔG: change in the free energy takes place while reactants convert to product where both are in standard state. It depends on the equilibrium constant K
ΔG =ΔGo+RTln(K)ΔGo=ΔHo−TΔSo
Where,
T is the temperature
ΔG is the free energy
ΔGo, ΔHo and ΔSo is standard free energy, enthalpy and entropy values.
1. For the four structures provided, Please answer the following questions in the table
below.
a. Please draw π molecular orbital diagram (use the polygon-and-circle method if
appropriate) and fill electrons in each molecular orbital
b. Please indicate the number of π electrons
c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non-
aromatic
TT MO diagram
Number of π e-
Aromaticity Evaluation (X choose one)
Non-aromatic
Aromatic
Anti-aromatic
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+
IV
1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?
Q3: Circle the molecules that are optically active:
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Chapter 20 Solutions
Chemistry: The Molecular Nature of Matter and Change (Looseleaf)
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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