Question

Calculate the change in Gibbs free energy for each of the
following sets of Δ*H*rxn ∘, Δ*S*∘rxn, and
*T*.

1) Δ*H*∘rxn=− 97 kJ , Δ*S*∘rxn=− 154 J/K ,
*T*= 310 K

Express your answer using two significant figures.

2) Δ*H*∘rxn=− 97 kJ , Δ*S*∘rxn=− 154 J/K ,
*T*= 856 K

3) Δ*H*∘rxn= 97 kJ , Δ*S*∘rxn=− 154 J/K ,
*T*= 310 K

4) Δ*H*∘rxn=− 97 kJ , Δ*S*∘rxn= 154 J/K ,
*T*= 408 K

Answer #1

Calculate the change in Gibbs free energy for each of the
following sets of ΔHrxn, ΔSrxn, and
T.
Part A
ΔH∘rxn= 90. kJ , ΔSrxn= 152 J/K , T=
303 K
Express your answer using two significant figures.
ΔG =
kJ
Part B
ΔH∘rxn= 90. kJ , ΔSrxn= 152 J/K , T=
750 K
Express your answer using two significant figures.
ΔG =
kJ
Part C
ΔH∘rxn= 90. kJ , ΔSrxn=− 152 J/K , T=
303 K
Express your answer...

Calculate the change in Gibbs free energy for each of the
following sets of ?H?rxn, ?S?rxn, and T.
Part A ?H?rxn=? 115 kJ ; ?S?rxn= 248 J/K ; T= 295 K
Part B ?H?rxn= 115 kJ ; ?S?rxn=? 248 J/K ; T= 295 K
Part C ?H?rxn=? 115 kJ ; ?S?rxn=? 248 J/K ; T= 295 K Express
your answer using two significant figures.
Part D ?H?rxn=? 115 kJ ; ?S?rxn=? 248 J/K ; T= 565 K Express
your answer...

A. Using given data, calculate the change in
Gibbs free energy for each of the following reactions. In each case
indicate whether the reaction is spontaneous at 298K under standard
conditions.
2H2O2(l)→2H2O(l)+O2(g)
Gibbs free energy for H2O2(l) is -120.4kJ/mol
Gibbs free energy for H2O(l) is -237.13kJ/mol
B. A certain reaction has ΔH∘ = + 35.4
kJ and ΔS∘ = 85.0 J/K . Calculate ΔG∘ for the
reaction at 298 K. Is the reaction spontaneous at
298K under standard
conditions?

he thermodynamic properties for a reaction are related by the
equation that defines the standard free energy, ΔG∘, in
kJ/mol:
ΔG∘=ΔH∘−TΔS∘
where ΔH∘ is the standard enthalpy change in kJ/mol and
ΔS∘ is the standard entropy change in J/(mol⋅K). A good
approximation of the free energy change at other temperatures,
ΔGT, can also be obtained by utilizing this
equation and assuming enthalpy (ΔH∘) and entropy
(ΔS∘) change little with temperature.
Part A
For the reaction of oxygen and nitrogen to...

1.
Calculate the standard free energy change at 500 K for the
following reaction.
Cu(s) +
H2O(g) à CuO(s) +
H2(g)
ΔH˚f
(kJ/mol)
S˚
(J/mol·K)
Cu(s)
0
33.3
H2O(g)
-241.8
188.7
CuO(s)
-155.2
43.5
H2(g)
0
130.6
2. When
solid ammonium nitrate dissolves in water, the resulting solution
becomes cold. Which is true and why?
a. ΔH˚
is positive and ΔS˚ is positive
b. ΔH˚
is positive and ΔS˚...

The chemical reaction that causes iron to corrode in air is
given by
4Fe+3O2→2Fe2O3
in which at 298 K
ΔH∘rxn
= −1684 kJ
ΔS∘rxn
= −543.7 J/K
Gibbs free energy (G) is a measure of the spontaneity of a
chemical reaction. It is the chemical potential for a reaction, and
is minimized at equilibrium. It is defined as G=H−TS where H is
enthalpy, T is temperature, and S is entropy.
Part A
What is the standard Gibbs free energy for...

Part A
Calculate the standard enthalpy change for the reaction
2A+B⇌2C+2D
where the heats of formation are given in the following
table:
Substance
ΔH∘f
(kJ/mol)
A
-227
B
-399
C
213
D
-503
Express your answer in kilojoules.
Answer= 273kJ
Part B:
For the reaction given in Part A, how much heat is absorbed when
3.70 mol of A reacts?
Express your answer numerically in kilojoules.
Part C:
For the reaction given in Part A, ΔS∘rxn is 25.0 J/K ....

Diamond
a. At 298 K, what is the Gibbs free energy change G for the
following reaction? Cgraphite -> Cdiamond
b. Is the diamond thermodynamically stable relative to graphite
at 298 K?
c. What is the change of Gibbs free energy of diamond when it is
compressed isothermally from 1 atm to 1000 atm at 298 K?
d. Assuming that graphite and diamond are incompressible,
calculate the pressure at which the two exist in equilibrium at 298
K.
e....

The chemical reaction that causes iron to corrode in air is
given by
4Fe+3O2→2Fe2O3
in which at 298 K
ΔH∘rxn
= −1684 kJ
ΔS∘rxn
= −543.7 J/K
Part A:
What is the standard Gibbs free energy for this reaction? Assume
the commonly used standard reference temperature of 298 K.
Part B:
What is the Gibbs free energy for this reaction at 3652 K ?
Assume that ΔH and ΔS do not change with
temperature.
Part C:
At what temperature Teq...

Given the values of ΔH∘rxn, ΔS∘rxn, and
T below, determine ΔSuniv.
Part A
ΔH∘rxn= 117 kJ , ΔS∘rxn=− 263 J/K ,
T= 291 K .
ΔSuniv=
−1.3•102
J/K
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Incorrect; Try Again; 3 attempts remaining
Part B
ΔH∘rxn=− 117 kJ , ΔS∘rxn= 263 J/K ,
T= 291 K .
ΔSuniv=
J/K
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Part C
ΔH∘rxn=− 117 kJ , ΔS∘rxn=− 263 J/K ,
T= 291 K.
ΔSuniv=
J/K
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Part D
ΔH∘rxn=− 117 kJ ,...

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