A graph of ln Pvap versus 1/T results in an equation of the best fit line being y = –3412.8x +16.9. What is the heat of vaporization?
a. 19.7 kJ/mole
b. 12.4 kJ/mole
c. 23.1 kJ/mole
d. 28.4 kJ/mole
URGENT CHEM HELP PLEASE!!!!!?
2020-07-10 8:59 pm
回答 (2)
2020-07-10 9:09 pm
Liquid ⇌ Vapor Kp = Pvap
ln(Kp) = (-ΔH/R)(1/T) + constant
Hence, ln(Pvap) = (-ΔH/R)(1/T) + constant
Plotting the graph of ln(Pvap) versus 1/T. a straight line is obtained with slope -ΔHvap/R.
Hence, -ΔHvap/R = -3412.8
-ΔHvap/8.314 = -3412.8
ΔHvap = 3412.8 × 8.314
Heat of vaporization. ΔHvap = 28400 J/mole = 28.4 kJ/mole
The answer: d. 28.4 kJ/mole
ln(Kp) = (-ΔH/R)(1/T) + constant
Hence, ln(Pvap) = (-ΔH/R)(1/T) + constant
Plotting the graph of ln(Pvap) versus 1/T. a straight line is obtained with slope -ΔHvap/R.
Hence, -ΔHvap/R = -3412.8
-ΔHvap/8.314 = -3412.8
ΔHvap = 3412.8 × 8.314
Heat of vaporization. ΔHvap = 28400 J/mole = 28.4 kJ/mole
The answer: d. 28.4 kJ/mole
2020-07-10 9:10 pm
Its the Clausius-Clapeyron equation
slope of this line = - ΔHvap/R
(-3412.8 K) = -ΔHvap/(8.314 J/Kmol)
ΔHvap = (3412.8 K)x (8.314 J/Kmol)
ΔHvap = 28.4 kJ/mol
slope of this line = - ΔHvap/R
(-3412.8 K) = -ΔHvap/(8.314 J/Kmol)
ΔHvap = (3412.8 K)x (8.314 J/Kmol)
ΔHvap = 28.4 kJ/mol
收錄日期: 2021-04-18 18:32:10
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