Answer:
ΔG = 6259.11 kJ/mol
Explanation:
At 298 K, the equilibrium vapor pressure of water above this solid is 14.7 Torr.
MSO4⋅5H2O(s)⇔ MSO4(s)+5H2O(g)
→ 1 atm = 760 torr
So, 14.7 torr= 0.0193 atm = 0.0193 / 760
Kp = [p(H2O)]3
= ( 0.0193/760)³
= 1.637 x 10^-14
as we know that, by Kp and Kc reaction
Kp = Kc (RT)^Δng
Δng = 3 -0 = 3
Kc = Kp / (RT)³
= 1.637 x 10^-14 / (0.082 x 298)³
= 1.637 x 10^-14 / 14,591.177
1.12237 x 10^-18
What is the value of Δ for the reaction when the vapor pressure of water is 14.7 Torr?
we have, Δ = - RT ln K
Δ = - (8.314 J/K-mol × 298) ln 1.12237 x 10^-18
= 2477.57 x ln 1.12237 x 10^-18
= 2477.57 x 2.5263
= 6259.11 kJ/mol
The value of ΔG for the reaction when the vapor pressure of water is 14.7 Torr is 6259.11 kJ/mol.
The ΔG is the change in free energy of a system that goes from initial state to final state.
Also known as Gibbs free energy.
Given,
At 298 K, the equilibrium vapor pressure of water above the solid is 14.7 Torr.
The reaction is
[tex]\rm MSO_4.3H_2O(s) <=>MSO_4(s)+3H_2O(g)[/tex]
1 atm = 760 torr
So, 14.7 torr = [tex]\dfrac{14.7}{760\;torr} = 0.0193\; atm.[/tex]
By the formula of the equilibrium constant
[tex]Kp = [p(H_2O)]^3 \\\\Kp= [0.0193]^3 =1.637 \times 10^-^1^4[/tex]
We know that Kp = Kc
[tex]Kp = Kc (RT)^\Delta^n^g[/tex]
Δng = 3 -0 = 3
[tex]Kp = \dfrac{ Kc}{(RT)^3}[/tex]
Putting the values
[tex]Kp = \dfrac{ 1.637 \times 10^-14 }{(0.082 \times 298)^3}\\\\Kp= \dfrac{ 1.637 \times 10^-14 }{14,591.177}\\\\ =1.12237 \times 10^-18[/tex]
Now, Δ = - RTlnK
[tex] \Delta = - (8.314 J/K-mol \times 298) ln 1.12237 \times 10^-^1^8 [/tex]
[tex] = 2477.57 \times ln 1.12237 \times 10^-^1^8 [/tex]
= 2477.57 x 2.5263
= 6259.11 kJ/mol
Thus, The value of ΔG for the reaction when the vapor pressure of water is 14.7 Torr is 6259.11 kJ/mol.
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