Answer:
The responses to the questions are:
a. The entropy of sublimation for CO2 is 134.07 J/Kmol.
b. The entropy for the universe in this reversible scenario is 376 J/K.
Explanation:
Entropy of sublimation refers to the change in entropy that occurs when one mole of a solid transitions to vapor at its sublimation point.
a. The mass of solid CO₂ is noted as 389 g.
The molar mass of CO₂ is 44.01 g/mol.
The number of moles of CO₂ in the sculpture is calculated as: Mass/(Molar mass)
= (389 g)/(44.01 g/mol) = 8.84 moles.
The entropy of sublimation is expressed as
ΔS =
- S
=
Where:
ΔH = 26.1 KJ/mol.
T = Temperature = –78.5°C = 194.65 K.
Consequently, the energy required for the 389 g of dry ice to sublime is: 26.1 KJ/mol × 8.84 moles = 230.695 KJ.
Subsequently, the entropy of sublimation is calculated as ΔS =
= 1.185 KJ/K
= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol.
b. The universe’s entropy can be computed as:
ΔS =
+ ΔS
If the heat absorbed by the system equals the heat released by the surroundings, then:
ΔS =
=1.185 KJ/K - = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K
= 376 J/K.
Response:
Here's my calculation
Clarification:
Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We need to determine the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's compile all the information in one location.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
3. Plot the initial values
The graph below visualizes the initial concentrations as plotted on the vertical axis.
