The equilibrium constant K p value should be derived given the equilibrium reaction at 300 ° C . Concept Information: In thermodynamics , free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work. The free energy is represented by the letter G. The equation given below helps us to calculate the change in standard free energy in a system. ΔG ° = Δ Η ° - T Δ S ° Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work. The free energy is represented by the letter G . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG ° rxn ) is the difference in free energy of the reactants and products in their standard state. ΔG ° =-RTln K ΔG = Free energy ΔG ° = Standard state free energy R = Gas Constant ( 0 .0826 l .atm/K .atm ) T = Temprature 273 K K= Equlibrium Constant (K P and K C ) ΔG ° rxn = ∑ nΔG f ° (Products)- ∑ nΔG f ° (Reactants)
The equilibrium constant K p value should be derived given the equilibrium reaction at 300 ° C . Concept Information: In thermodynamics , free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work. The free energy is represented by the letter G. The equation given below helps us to calculate the change in standard free energy in a system. ΔG ° = Δ Η ° - T Δ S ° Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work. The free energy is represented by the letter G . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG ° rxn ) is the difference in free energy of the reactants and products in their standard state. ΔG ° =-RTln K ΔG = Free energy ΔG ° = Standard state free energy R = Gas Constant ( 0 .0826 l .atm/K .atm ) T = Temprature 273 K K= Equlibrium Constant (K P and K C ) ΔG ° rxn = ∑ nΔG f ° (Products)- ∑ nΔG f ° (Reactants)
Solution Summary: The author explains that free energy is used to explain the total energy content in a thermodynamic system that can be converted into work.
Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.
Chapter 17, Problem 17.84QP
Interpretation Introduction
Interpretation:
The equilibrium constant Kp value should be derived given the equilibrium reaction at 300°C.
Concept Information:
In thermodynamics, free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work. The free energy is represented by the letter G.
The equation given below helps us to calculate the change in standard free energy in a system.
ΔG° = ΔΗ°- TΔS°
Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work. The free energy is represented by the letter G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.
ΔG°=-RTlnK ΔG=Free energyΔG°=Standard state free energyR=GasConstant(0.0826l.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)
5
What would the complete ionic reaction be if aqueous solutions of potassium sulfate and barium acetate were mixed?
ed
of
Select one:
O a
2 K SO4 + Ba2 +2 C₂H3O21
K+SO4 + Ba2+ + 2 C2H3O21
K+SO42 + Ba2 +2 C2H3O2
BaSO4 +2 K+ + 2 C2H3O
estion
Ob.
O c.
Od.
2 K SO4 +Ba2 +2 C₂H₂O₂
BaSO4 + K+ + 2 C2H3O
BaSO4 + K + 2 C2H301
→Ba² +SO42 +2 KC2H3O
s page
(28 pts.) 7. Propose a synthesis for each of the following transformations. You must include the
reagents and product(s) for each step to receive full credit. The number of steps is provided.
(OC 4)
4 steps
4 steps
OH
b.
LTS
Solid:
AT=Te-Ti
Trial 1
Trial 2
Trial 3
Average
ΔΗ
Mass water, g
24.096
23.976
23.975
Moles of solid, mol
0.01763
001767
0101781
Temp. change, °C
2.9°C
11700
2.0°C
Heat of reaction, J
-292.37J -170.473
-193.26J
AH, kJ/mole
16.58K 9.647 kJ 10.85 kr
16.58K59.64701
KJ
mol
12.35k
Minimum AS,
J/mol K
41.582
mol-k
Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn &
Show your calculations for:
AH in J and then in kJ/mole for Trial 1:
qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J
qsin =
qrxn =
292.35 292.37J
AH in J = 292.375 0.2923kJ
0.01763m01
=1.65×107
AH in kJ/mol =
=
16.58K
0.01763mol
mol
qrx
Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con
Kelvin.)
AS=AHIT
(1.65×10(9.64×103) + (1.0
Jimai
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