Answer:
See explanation and image attached
Explanation:
The compound SF4 is sulphur tetraflouride. It has a zero formal charge and has a total of 34 valence electrons.
The central atom in the molecule is sulphur in an sp3d hybridization state hence the molecule has a trigonal bipyramidal electron geometry. Since it is a molecule of the sort AX4E; it gives a see-saw molecular shape.
The structure of the molecule is shown in the image attached to this answer.
The question is incomplete, complete question is :
Consider the decomposition of red mercury(II) oxide under standard state conditions.
Given :
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:
Given :
Entropy change of the reaction ; ΔS
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.