In: Chemistry
For the reaction CuS(s) + H2(g) H2S(g) + Cu(s) ΔG°f (CuS) = −53.6 kJ/mol ΔG°f (H2S) = −33.6 kJ/mol ΔH°f (CuS) = −53.1 kJ/mol ΔH°f (H2S) = − 20.6 kJ/mol a. Calculate ΔG° and ΔH° at 298 K and 1 atm pressure. b. Will this reaction proceed spontaneously at 298 K and 1 atm pressure? c. Calculate the equilibrium constant for this reaction at 298 K. d. Calculate ΔS° at 298 K and 1 atm pressure. e. Calculate ΔG at 798 K and 1 atm pressure (assume ΔS° and ΔH° do not change with temperature). f. Calculate the equilibrium constant at 798 K and 1 atm pressure.
a.
To calculate Go
(reaction)
Go
(reaction) = (sum of standard change in free energies of formation
of products) - (sum of standard change in free energies of
formation of reactants)
Go
(reaction) =
ai
Gofproducts
-
bi
Gofreactants
Also standard change in free energies of formation for elements is zero.
Therefore, Go
(reaction) =
Gof(H2S)
-
Gof
(CuS)
Substituting the given values, Go
(reaction) = -33.6 - (-53.6) = +20.0 kJ/mol
Therefore Go
(reaction) = +20.0 kJ/mol
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To calculate Ho
(reaction)
Ho
(reaction) =
ai
Hofproducts
-
bi
Hofreactants
Also standard change in enthalpies of formation for elements is zero.
Therefore, Ho
(reaction) =
Hof(H2S)
-
Hof
(CuS)
Substituting the given values, Ho
(reaction) = -20.6 - (-53.1) = +32.5 kJ/mol
Therefore, Ho
(reaction) = +32.5 kJ/mol
Answer for a. is Go
(reaction) = +20.0 kJ/mol and
Ho
(reaction) = +32.5 kJ/mol
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b.
298K and 1 atm pressure are standard conditions and at standard
conditions change in free energy G of reaction is
nothing but standard change in free energy
Go of
reaction.
For a reaction to be spontaneous change in free energy must be negative but we see that
G (reaction) =
Go
(reaction) = +20.0 kJ/mol , is positive at 298K and 1 atm
pressure.
Hence, the reaction will NOT proceed spontaneously at 298K and 1 atm pressure.
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c.
Relation between standard change in free energy of reaction
Go
(reaction) and equilibrium constant K of the reaction is given by
equation,
Go
(reaction) = -RTlnK
R = universal gas constant = 8.314 J K-1mol-1 ; T = absolute temperature = 298 K (given)
We have calculated Go
(reaction) to be +20.0 kJ/mol = 20*103 J/mol
Substituting the values in equation we get,
Go
(reaction) = -RTlnK
20*103 = - 8.314 * 298* lnK
lnK = -8.0724 but lnK = 2.303 log K ( since lnx = 2.303logx)
Therefore, 2.303 log K = -8.0724 or logK = -8.0724 / 2.303 = -3.5052
logK = -3.5052 = -3 -0.5052 = -3 - 0.5052 - 1 +1 = (-3-1) + (1-0.5052) = - 4 + 0.4948
logK = - 4 + 0.4948
K = 10-4 * antilog 0.4948 = 10-4 *3.125
Therefore equilibrium constant K = 3.125 * 10-4 at 298K
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d.
Standard change in entropy So can
be calculated using the relation,
Go =
Ho -
T
So
So, So =
(
Ho -
Go ) /
T
We have, Go =
+20.0 kJ/mol,
Ho =
+32.5 kJ/mol and T = 298 K
Substituting the values we get, So =
(32.5 - 20) / 298 = 0.04195 kJ mol-1K-1 = +
41.95 J mol-1K-1
So = +
41.95 J mol-1K-1 at
298K.
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e.
Change in free energy of reaction G is given by
equation,
G =
H -
T
S
Given, T = 798 K,
Assuming H and
S
are same as
Ho and
So
H =
Ho =
+32.5 kJ/mol = +32.5 * 103 J/mol and
S =
So = +
41.95 J mol-1K-1
Substituting the values we get,
G =
H -
T
S = (32.5 *
103) - (798 * 41.95 ) = - 976.1 J/mol
Therefore, at 798 K, G = - 976.1 J/mol
= 0.976 kJ/mol
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f.
Given, T = 798 K
Equilibrium constant K can be calculated using the equation,
Go
(reaction) = -RTlnK
Go
(reaction) = +20.0 kJ/mol = 20*103
J/mol ( as calculated under question a)
R = universal gas constant = 8.314 J K-1mol-1 ; T = absolute temperature = 798 K (given)
Substituting the values we get,
Go
(reaction) = -RTlnK
20*103 = - 8.314 * 798 * lnK
20*103 = - 8.314 * 798 * 2.303 log K (since lnx = 2.303logx)
log K = - (20*103) / (8.314 * 798 * 2.303) = -1.3089
log K = -1.3089 = -1 - 0.3089 = -1 -0.3089 +1 -1 = (-1 -1) + (+1 - 0.3089) = - 2 + 0.6911
log K = - 2 + 0.6911
K = 10-2 * antilog 0.6911 = 10-2 *4.910
Therefore equilibrium constant K = 4.91 * 10-2 at 798 K