Question

Calculate the **standard molar Gibbs free energy change
(****DeltaG _{rxn}^{0})**

3 C_{2}H_{4}(g) ®
C_{6}H_{6}(g) + 3 H_{2}(g).

Answer #1

Hf = Heat of formation and Sf = Entropy of formation

C2H4 : Hf = 52.26 kJ/mol, Sf = 219.56 J/K.mol and Gf = 68.15 kJ/mol (at 298 K)

C6H6 : Hf = 49 kJJ/mol ; Sf = 173.3 J/K.mol ; Gf = 124.3 kJ/mol (298 K)

H2 : Hf = 0 ; Gf = 0 ; Sf = 130.68 J/K.mol

Delta Hf of reaction = Hf C6H6 - 3* Hf C2H4 = -107.75 kJ/mol

Delta Sf of reaction = Sf C6H6 + 3*Sf H2 - 3*Sf C2H4 = -93.34 J/K.mol

Delta G = Delta Hf - T * delta Sf

At 298 K : Delta G = -79.73 kJ/mol

At 400 K : delta G = -70.4 kJ/mol : Here we have assumed that Hf and Sf do not vary with temperature.

Verify the data for 298 K by using the Gf values at 298 K for 3 gases.

Delta Gf 298 K = Gf C6H6 - 3* C2H4 = -80.15 kJ/mol which is approx same as calculated.

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....

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˚...

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?

For a gaseous reaction, standard conditions are 298 K and a
partial pressure of 1 atm for all species.
For the reaction
C2H6(g)+H2(g)↽−−⇀2CH4(g)
the standard change in Gibbs free energy is Δ?°=−69.0 kJ/mol.
What is ΔG for this reaction at 298 K when the partial
pressures are ?C2H6=0.300 atm, ?H2=0.500 atm, and ?CH4=0.950
atm?

The standard molar entropy of benzene is 173.3 J/K-mol.
Calculate the change in its standard molar Gibbs energy when
benzene is heated from 25C to 45C.

Given the following information,
calculate the standard Gibbs free energy of the reaction at 1455 K.
State if the reaction is spontaneous or nonspontaneous followed by
the temperature at which the reaction switches spontaneity if
applicable.
CaCO3(s) --> CaO(s) + CO2(g)
Given: Δ°H = 179.2 kJ , Δ°S = 160.2 J/K

Calculate the change in the molar Gibbs energy of hydrogen gas
when its pressure is increased isothermally from 1.0 atm to 100.0
atm at 298 K.

Calculate the standard Helmholtz energy of formation, ΔfA, of
CH3OH(l) at 298 K from the standard Gibbs energy of formation and
the assumption that H2 and O2 are perfect gases.

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....

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