Respuesta :
The question is incomplete, complete question is :
Consider the decomposition of red mercury(II) oxide under standard state conditions.
[tex]2HgO(s)\rightarrow 2Hg(l)+O_2(g)[/tex]
Given :
[tex]\Delta S_{HgO}^o=70.29 J/mol K[/tex]
[tex]\Delta S_{Hg}^o=75.9 J/mol K[/tex]
[tex]\Delta S_{O_2}^o=205.2 J/mol K[/tex]
Enthalpy change of the reaction = ΔH = -90.83 kJ/mol
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
Answer:
a) The decomposition is spontaneous under standard state conditions.
b)The reaction will spontaneous above -419.69 Kelvins.
Explanation:
[tex]2HgO(s)\rightarrow 2Hg(l)+O_2(g)[/tex]
Given :
[tex]\Delta S_{HgO}^o=70.29 J/mol K[/tex]
[tex]\Delta S_{Hg}^o=75.9 J/mol K[/tex]
[tex]\Delta S_{O_2}^o=205.2 J/mol K[/tex]
Entropy change of the reaction ; ΔS
[tex]\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o][/tex]
[tex]=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K[/tex]
Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K
1 kJ = 1000 J
At standard condition the value of temperature = T = 298 K
ΔG = ΔH - TΔS
ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol
ΔG < 0 ( spontaneous)
The decomposition is spontaneous under standard state conditions.
b) Above what temperature does the reaction become spontaneous
Let the ΔG = 0
Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K
ΔG = ΔH - TΔS
0 = -90830 J/mol K - T × 216.42 J/mol K
T = -419.69 K
The reaction will spontaneous above -419.69 Kelvins.
