Answer:
a) ΔHvap=35.3395 kJ/mol
b) Tb=98.62 °C
Explanation:
Given the reaction:
C₇H₁₆ (l) ⇔ C₇H₁₆ (g)
Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.
When
T₁=50°C=323.15K ⇒P₁=0.179
T₂=86°C=359.15K ⇒P₂=0.669
The Clasius-Clapeyron equation is:
ΔHvap=35339.5 J/mol=35.3395 KJ/mol
Normal boiling point ⇒ P=1 atm
Hence, we find the normal boiling point where:
T₁=323.15K
P₁=0.179 atm
P₂=1 atm
T₂=371.77 K= 98.62 °C
482VP I think is the correct answer.
The net amount of energy produced can be obtained from a table of enthalpy change of formation, available online.
The enthalpy change of formation indicate how much energy the 1 mole of the product (H2O) has relative to the elemental reactants (H2 and O2). In other words, the "lost" energy equals the heat/energy released.
For water (H2O), this value is -285.8 if the final product is a liquid under standard conditions, and -241.82 if the product is in gas form which contains some energy that could be further released. This means that if the final product (H2O) is in liquid form, energy released is 285.8 kJ/mol.
Since water is in liquid form under standard conditions, the first value (285.8 kJ/mol) is generally appropriate.
Yes. because equation balances the number of particles.
STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P
V2/V1= T2*P1/T1*P2
V2/V1=273.15K* 90^10^3Pa/ 300.15K * 1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml
