Question

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 Δ*G* for this reaction at 298 K when the partial
pressures are ?C2H6=0.300 atm, ?H2=0.500 atm, and ?CH4=0.950
atm?

Answer #1

The equation for Gibbs Free Energy is

G = + RT lnQ

For a gas phase reaction

aA(g) + bB(g) cC(g) + dD(g)

Q_{p} =

hence

For Given Reaction

where
= -69.0kJ/mol ; T = 298K ; R = 8.314 J K^{-1}
mol^{-1}

PC2H6 = 0.300 atm ; PH2 = 0.500 atm ; PCH4 = 0.950 atm

= -69000 J mol^{-1} + ( 8.314 J K^{-1}
mol^{-1} )(298 K)

= -69000 J mol^{-1} + 4446.086 J mol^{-1}

^{
=} - 64553.914 J mol^{-1}

= - 64.55 kJ mol^{-1}

For a gaseous reaction, standard conditions are 298 K and a
partial pressure of 1 atm for all species. For the reaction
N2(g) + 3H2(g) <--> 2NH3 (g)
the standard change in Gibbs free energy is ΔG° = -69.0 kJ/mol.
What is ΔG for this reaction at 298 K when the partial pressures
are
PN2= 0.250 atm, PH2 = 0.450 atm, and PNH3 = 0.800 atm

In Part A, we saw that ΔG∘=−242.1 kJ for the
hydrogenation of acetylene under standard conditions (all pressures
equal to 1 atm and the common reference temperature 298 K). In Part
B, you will determine the ΔG for the reaction under a
given set of nonstandard conditions.
Part B
At 25 ∘C the reaction from Part A has a composition as shown in
the table below.
Substance
Pressure
(atm)
C2H2(g)
3.95
H2(g)
5.65
C2H6(g)
5.25×10−2
What is the free energy...

In Part A, we saw that ΔG∘=−242.1 kJ for the
hydrogenation of acetylene under standard conditions (all pressures
equal to 1 atm and the common reference temperature 298 K). In Part
B, you will determine the ΔG for the reaction under a
given set of nonstandard conditions.
At 25 ∘C the reaction from Part A has a composition as shown in
the table below.
Substance
Pressure
(atm)
C2H2(g)
5.35
H2(g)
3.95
C2H6(g)
4.25×10−2
What is the free energy change, ΔG,...

Acetylene, C2H2, can be converted to ethane, C2H6, by a process
known as hydrogenation. The reaction is
C2H2(g)+2H2(g)⇌C2H6(g)
Given the following data at standard conditions (all pressures
equal to 1 atm and the common reference temperature 298 K), what is
the value of Kp for this reaction?
Substance
ΔG∘f
(kJ/mol)
C2H2(g)
209.2
H2(g)
0
C2H6(g)
−32.89
Express your answer using two significant figures.

Consider the following reaction starting with standard state
conditions at 298 K:
N2O4(g)⇌2NO2(g) ΔG∘rxn=5.4kJ
Determine the partial pressures of the N2O4
and NO2 at the start of the reaction. In other
words, what are each of their initial standard state
conditions?
The question is asking for only one value. Please show steps on
how to solve.

Consider the reaction 2NO2(g)→N2O4(g)
Calculate ΔG at 298 K if the partial pressures of NO2 and N2O4
are 0.37 atm and 1.62 atm , respectively

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....

Acetylene, C2H2, can be converted to ethane, C2H6, by a process
known as hydrogenation. The reaction is
C2H2(g)+2H2(g)⇌C2H6(g)
Given the following data at standard conditions (all pressures
equal to 1 atm and the common reference temperature 298 K), what is
the value of Kp for this reaction?
Substaance .. G∘f (kJ/mol)
C2H2(g)
209.2
H2(g) 0
C2H6(g)
-32.89

Calculate the work (w) and ΔEo, in kJ, at 298 K and 1
atm pressure, for the combustion of one mole of
C4H10 (g). First write and balance the
equation. The products will be CO2 (g) and
H2O (g).
The value of ΔHo for this reaction is -2658.3
kJ/mol.
The value for w in kJ =
The value for ΔEo ( in kJ) =

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