The heat of vaporization of water at 100°C is 40.66 kJ/mol. Calculate the quantity of heat that is absorbed/re?

Question

1) The heat of vaporization of water at 100°C is 40.66 kJ/mol. Calculate the quantity of heat that is absorbed or released when 9.00 g of steam condenses to liquid water at 100°C.
2) Calculate the total quantity of heat required to convert 25.0 g of liquid CCl4(l) from 35.0°C to gaseous CCl4 at 76.8°C (the normal boiling point for CCl4). The specific heat of CCl4(l) is 0.857 J/(g·°C), its heat of fusion is 3.27 kJ/mol, and its heat of vaporization is 29.82 kJ/mol.
3) Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm Hg at 35°C and has a ΔHvap of 32.1 kJ/mol.
4) The enthalpy change for converting 1.00 mol of ice at -50.0°C to water at 70.0°C is _____ kJ. The specific heats of ice, water, and steam are 2.09 J/(g·K), 4.18 J/(g·K), and 1.84 J/(g·K), respectively. For H2O, the heat of fusion (Hfus) is 6.01 kJ/mol, and the heat of vaporization (Hvap) is 40.67 kJ/mol.
If you could help me with any of these, I would greatly appreciate it! Thank you!

Anonymous · Accepted Answer

1. Q = moles steam x molar heat of vaporization

Q = 9g / 18g/mole x 40.66kJ/mole = 20.33kJ

2. this requires the heat to warm liquid CCl4 from 25 to 76.8ºC and then to vaporize all CCl4 at this temperature

Q1 = mass CCl4 x specific heat CCl4 x ∆T = J

Q2 = moles CCl4 x molar heat of vaporization = kJ

watch that you convert J to kJ or kJ to J. in other words, watch your units

3. Claussius-Clapeyron equation

lnPº = (-∆Hvap / R)(1/T2 – 1/Tbpº)

convert mmHg to atm 55.1mmHg / 760mmHg/atm = 0.0725atm

ln(0.0725atm) = (-32100J/mole/0.0821L-atm/mole-K)(1/309 – 1/Tbp)

-2.624 = -3.91×10^5K(0.00324 – 1/Tbp)

6.71×10^-6 = 0.00324 – 1/Tbp

3.23×10^-3 = 1/Tbp

309.28K = bp = 36.28ºC

4. Q1 = heating ice from -50ºC to 0ºC = mass x specific heat ice x ∆T1

Q2 = melting ice at 0ºC = moles ice x molar heat of fusion

Q3 = heating water from 0ºC to 70ºC = mass water x specific heat water x ∆T2

add Q1, Q2, and Q3, watch the units

Anonymous · Answer

actual not a hundred% particular the thank you to do this one. i needed to declare I made a correction on different situation and wanted you to be conscious of. I appeared up tables of vapor stress of water and the respond is between 31 T at 30C and fifty 5 T at 40C so 5) appears like the marvelous answer. yet unsure on the thank you to derive.

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The heat of vaporization of water at 100°C is 40.66 kJ/mol. Calculate the quantity of heat that is absorbed/re?

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