It should be explained that whether the sign of entropy change for system, surroundings and universe is greater than zero or not. Concept introduction: The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings. ΔS o ( universe ) = ΔS o ( system ) +ΔS o ( surroundings ) The ΔS o ( universe ) should be greater than zero for a spontaneous process. The ΔS o ( system ) can be calculated by the following expression, ΔS o ( system ) = Δ r S ° = ∑ nS ° ( products ) - ∑ nS ° ( reactants ) The ΔS o ( surroundings ) can be calculated by the following expression, ΔS o ( surroundings ) = - Δ r H o T Here, Δ r H ° is the enthalpy change for the reaction. The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔG o . It is related to entropy and entropy by the following expression, Δ r G o =Δ r H o -TΔ r S o ΔG o is also related to the equilibrium constant K by the equation, Δ r G o = -RTlnK p The rearranged expression is, K p = e - Δ r G o RT
It should be explained that whether the sign of entropy change for system, surroundings and universe is greater than zero or not. Concept introduction: The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings. ΔS o ( universe ) = ΔS o ( system ) +ΔS o ( surroundings ) The ΔS o ( universe ) should be greater than zero for a spontaneous process. The ΔS o ( system ) can be calculated by the following expression, ΔS o ( system ) = Δ r S ° = ∑ nS ° ( products ) - ∑ nS ° ( reactants ) The ΔS o ( surroundings ) can be calculated by the following expression, ΔS o ( surroundings ) = - Δ r H o T Here, Δ r H ° is the enthalpy change for the reaction. The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔG o . It is related to entropy and entropy by the following expression, Δ r G o =Δ r H o -TΔ r S o ΔG o is also related to the equilibrium constant K by the equation, Δ r G o = -RTlnK p The rearranged expression is, K p = e - Δ r G o RT
Solution Summary: The author explains that the sign of entropy change for system, surroundings and universe is greater than zero or not.
It should be explained that whether the sign of entropy change for system, surroundings and universe is greater than zero or not.
Concept introduction:
The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.
ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)
The ΔSo(universe) should be greater than zero for a spontaneous process.
The ΔSo(system) can be calculated by the following expression,
ΔSo(system)= ΔrS°= ∑nS°(products)-∑nS°(reactants)
The ΔSo(surroundings) can be calculated by the following expression,
ΔSo(surroundings)=- ΔrHoT
Here, ΔrH° is the enthalpy change for the reaction.
The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,
ΔrGo=ΔrHo-TΔrSo
ΔGo is also related to the equilibrium constant K by the equation,
ΔrGo= -RTlnKp
The rearranged expression is,
Kp= e- ΔrGoRT
(b)
Interpretation Introduction
Interpretation:
The sign of ΔrH° and ΔrGo value for the given reaction should be predicted.
Concept introduction:
The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.
ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)
The ΔSo(universe) should be greater than zero for a spontaneous process.
The ΔSo(system) can be calculated by the following expression,
ΔSo(system)= ΔrS°= ∑nS°(products)-∑nS°(reactants)
The ΔSo(surroundings) can be calculated by the following expression,
ΔSo(surroundings)=- ΔrHoT
Here, ΔrH° is the enthalpy change for the reaction.
The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,
ΔrGo=ΔrHo-TΔrSo
ΔGo is also related to the equilibrium constant K by the equation,
ΔrGo= -RTlnKp
The rearranged expression is,
Kp= e- ΔrGoRT
(c)
Interpretation Introduction
Interpretation:
It should be identified that whether the value of Kp is very large, very small or nearer to 1 also that the equilibrium constant will be larger or smaller for temperatures greater than 298K.
Concept introduction:
The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.
ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)
The ΔSo(universe) should be greater than zero for a spontaneous process.
The ΔSo(system) can be calculated by the following expression,
ΔSo(system)= ΔrS°= ∑nS°(products)-∑nS°(reactants)
The ΔSo(surroundings) can be calculated by the following expression,
ΔSo(surroundings)=- ΔrHoT
Here, ΔrH° is the enthalpy change for the reaction.
The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,
ΔrGo=ΔrHo-TΔrSo
ΔGo is also related to the equilibrium constant K by the equation,
An orange laser has a wavelength of 610 nm. What is the energy of this light?
The molar absorptivity of a protein in water at 280 nm can be estimated within ~5-10% from its content of the amino acids tyrosine and tryptophan and from the number of disulfide linkages (R-S-S-R) between cysteine residues:
Ε280 nm (M-1 cm-1) ≈ 5500 nTrp + 1490 nTyr + 125 nS-S
where nTrp is the number of tryptophans, nTyr is the number of tyrosines, and nS-S is the number of disulfide linkages. The protein human serum transferrin has 678 amino acids including 8 tryptophans, 26 tyrosines, and 19 disulfide linkages. The molecular mass of the most dominant for is 79550.
Predict the molar absorptivity of transferrin.
Predict the absorbance of a solution that’s 1.000 g/L transferrin in a 1.000-cm-pathlength cuvet.
Estimate the g/L of a transferrin solution with an absorbance of 1.50 at 280 nm.
In GC, what order will the following molecules elute from the column?
CH3OCH3, CH3CH2OH, C3H8, C4H10
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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