It is given that, a perfectly insulated vessel 1 contains an ice cube at 0 ° C and water at 0 ° C , whereas perfectly insulated vessel 2 contains an ice cube at 0 ° C and saltwater solution at 0 ° C . The sign of Δ S univ and Δ S and Δ S surr is to be predicted for each, vessel 1 and 2 for the given process. Concept introduction: In a perfectly insulated system, no heat is lost. The value of Δ S surr for such system will be zero. To predict: The sign of Δ S univ , Δ S and Δ S surr in vessel 1 for the process H 2 O ( s ) → H 2 O ( l ) .
It is given that, a perfectly insulated vessel 1 contains an ice cube at 0 ° C and water at 0 ° C , whereas perfectly insulated vessel 2 contains an ice cube at 0 ° C and saltwater solution at 0 ° C . The sign of Δ S univ and Δ S and Δ S surr is to be predicted for each, vessel 1 and 2 for the given process. Concept introduction: In a perfectly insulated system, no heat is lost. The value of Δ S surr for such system will be zero. To predict: The sign of Δ S univ , Δ S and Δ S surr in vessel 1 for the process H 2 O ( s ) → H 2 O ( l ) .
Solution Summary: The author explains that a perfectly insulated vessel 1 contains an ice cube and water at 0°
Interpretation: It is given that, a perfectly insulated vessel
1 contains an ice cube at
0°C and water at
0°C, whereas perfectly insulated vessel
2 contains an ice cube at
0°C and saltwater solution at
0°C. The sign of
ΔSuniv and
ΔS and
ΔSsurr is to be predicted for each, vessel
1 and
2 for the given process.
Concept introduction: In a perfectly insulated system, no heat is lost. The value of
ΔSsurr for such system will be zero.
To predict: The sign of
ΔSuniv,ΔS and
ΔSsurr in vessel
1 for the process
H2O(s)→H2O(l).
(b)
Interpretation Introduction
Interpretation: It is given that, a perfectly insulated vessel
1 contains an ice cube at
0°C and water at
0°C, whereas perfectly insulated vessel
2 contains an ice cube at
0°C and saltwater solution at
0°C. The sign of
ΔSuniv and
ΔS and
ΔSsurr is to be predicted for each, vessel
1 and
2 for the given process.
Concept introduction: In a perfectly insulated system, no heat is lost. The value of
ΔSsurr for such system will be zero.
To predict: The signs of
ΔSuniv,ΔS and
ΔSsurr in vessel
2 for the process
H2O(s)→H2O(l) .
need help please and thanks dont understand only need help with C-F
Learning Goal:
As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7).
Part A - Difference in binding free eenergies
Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.
The margin of error is 2%.
Part B - Compare difference in free energy to the thermal…
Please correct answer and don't used hand raiting
need help please and thanks dont understand a-b
Learning Goal:
As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7).
Part A - Difference in binding free eenergies
Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.
The margin of error is 2%.
Part B - Compare difference in free energy to the thermal energy
Divide the…
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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