Question

Start from the following expression that relates the equilibrium constant K to the standard free energy (on a molecular basis)

We know that , where ∆*H ^{0}* and
∆

From the above information, find an expression for the slope of
ln(*K*) when plotted against (1/*T*). In other words,
find an expression for

Answer #1

Solution :

the following expression that relates the equilibrium constant K to the standard free energy :

∆Go = -RT lnK...........................1

where ∆Go = standard free energy

We know that ∆Go = ∆Ho - T ∆So .........................2

where ∆*H0* and ∆*S0* are the standard enthalpy
and entropy.

From equation 1 and 2,

∆Ho - T ∆So = -RT lnK

lnK = -(∆Ho - T ∆So) / (RT)

lnK = -∆Ho/ RT + ∆So/R

Com pare this equation with y = mx + c,

y = lnK,

x = 1/T,

Now if we draw a line of y vs x we get,

Slope of line = m = -∆Ho/ R

Please give feedback.

� Gibbs Free Energy: Equilibrium Constant
Nitric oxide, NO, also known as nitrogen monoxide, is one of the
primary contributors to air pollution, acid rain, and the depletion
of the ozone layer. The reaction of oxygen and nitrogen to form
nitric oxide in an automobile engine is
N2(g)+O2(g)?2NO(g)
The spontaneity of a reaction can be determined from the free
energy change for the reaction, ?G?.
A reaction is spontaneous when the free energy change is less
than zero.
A reaction...

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.
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Acetylene,...

What is the standard free energy change and equilibrium constant
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2Ag+ (aq) + Fe (s) 2 Ag (s) +
Fe2+ (aq)
Given standard electrode potentials:
Ag+ (aq) + e- Ag (s)
E0 = 0.7996 V
Fe2+ (aq) + 2 e- Fe (s)
E0 = - 0.447 V

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.
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1.
Calculate the standard free energy change at 500 K for the
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Cu(s) +
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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
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b. ΔH˚
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K.
e....

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