Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 1.1 g of octane is mi
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Answer:
The answers to the questions are;
a. The entropy of sublimation for carbon dioxide (the system) is
134.07 J/Kmol.
b. The entropy of the universe for this reversible process is 376 J/K.
Explanation:
Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at the temperature where the sublimation is taking place
a. We note that the mass of the solid CO₂ = 389 g
Molar mass of CO₂ = 44.01 g/mol
Number of moles of CO₂ in the sculpture = Mass/(Molar mass)
= (389 g)/(44.01 g/mol) = 8.84 Moles
Entropy of sublimation is given by
ΔS =
- S
=
Where:
ΔH = 26.1 KJ/mol
T = Temperature = –78.5°C = 194.65 K
Therefore the amount of heat required to cause the 389 g of dry ice to sublime = 26.1 KJ/mol × 8.84 Moles = 230.695 KJ
Therefore the entropy of sublimation = ΔS =
= 1.185 KJ/K
= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol
b. The entropy of the universe is given by;
ΔS =
+ ΔS
If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;
ΔS =
=1.185 KJ/K - = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K
= 376 J/K.