Question

he thermodynamic properties for a reaction are related by the
equation that defines the standard free energy, Δ*G*∘, in
kJ/mol:

Δ*G*∘=Δ*H*∘−*T*Δ*S*∘

where Δ*H*∘ is the standard enthalpy change in kJ/mol and
Δ*S*∘ is the standard entropy change in J/(mol⋅K). A good
approximation of the free energy change at other temperatures,
Δ*G**T*, can also be obtained by utilizing this
equation and assuming enthalpy (Δ*H*∘) and entropy
(Δ*S*∘) change little with temperature.

**Part A**

For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.):

ΔH∘rxn |
180.5kJ/mol |

ΔS∘rxn |
24.80J/(mol⋅K) |

Calculate the temperature in kelvins above which this reaction is spontaneous.

Express your answer to four significant figures and include the appropriate units.

The standard free energy change, Δ*G*∘, and the
equilibrium constant *K* for a reaction can be related by
the following equation:

Δ*G*∘=−*R**T*ln*K*

where *T* is the Kelvin temperature and *R* is
equal to 8.314 J/(mol⋅K).

**Part B**

The thermodynamic values from part A will be useful as you work through part B:

ΔH∘rxn |
180.5kJ/mol |

ΔS∘rxn |
24.80J/(mol⋅K) |

Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C:

N2(g)+O2(g)→2NO(g)

Express your answer numerically to three significant figures.

Answer #1

A)

ΔHo = 180.5 KJ/mol

ΔSo = 24.8 J/mol.K

= 0.0248 KJ/mol.K

use:

ΔGo = ΔHo - T*ΔSo

for reaction to be spontaneous, ΔGo should be negative

that is ΔGo<0

since ΔGo = ΔHo - T*ΔSo

so, ΔHo - T*ΔSo < 0

180.5- T *0.0248 < 0

T *0.0248 > 180.5

T > 7278 K

Answer: 7278 K

B)

ΔHo = 180.5 KJ/mol

ΔSo = 24.8 J/mol.K

= 0.0248 KJ/mol.K

T= 25.0 oC

= (25.0+273) K

= 298 K

use:

ΔGo = ΔHo - T*ΔSo

ΔGo = 180.5 - 298.0 * 0.0248

ΔGo = 173.1096 KJ/mol

T = 298 K

ΔG = 173.1096 KJ/mol

ΔG = 173109.6 J/mol

use:

ΔG = -R*T*ln Kc

173109.6 = - 8.314*298.0* ln(Kc)

ln Kc = -69.8707

Kc = 4.524*10^-31

Answer: 4.52*10^-31

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 ....

The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG∘, using the
following equation:
ΔG∘=−RTlnK
where T is a specified temperature in kelvins (usually
298 K) and R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values....

The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG∘, using the
following equation:
ΔG∘=−RTlnK
where T is a specified temperature in kelvins (usually
298 K) and R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values....

± Free Energy and Chemical Equilibrium
The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG, using the
following equation:
ΔG∘=−RTlnK
where T is standard temperature in kelvins and
R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the...

Item 5
The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG, using the
following equation:
ΔG∘=−RTlnK
where T is standard temperature in kelvins and
R is the gas constant.
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values.
Part A
Acetylene,...

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˚...

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 ΔHrxn ∘, ΔS∘rxn, and
T.
1) ΔH∘rxn=− 97 kJ , ΔS∘rxn=− 154 J/K ,
T= 310 K
Express your answer using two significant figures.
2) ΔH∘rxn=− 97 kJ , ΔS∘rxn=− 154 J/K ,
T= 856 K
3) ΔH∘rxn= 97 kJ , ΔS∘rxn=− 154 J/K ,
T= 310 K
4) ΔH∘rxn=− 97 kJ , ΔS∘rxn= 154 J/K ,
T= 408 K

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...

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