Exercises
Exercise 1. Calculate the amount of heat required to convert 25.0 g of benzene (C6H6) from 22.0°C to 90.0°C.
Melting point of benzene = 5.5°C
ΔHfus = 9.87 kJ/mol
ΔHvap = 33.2 kJ/mol
\(\displaystyle c(s)=1.05\;\frac{J}{g⋅K}\)
\(\displaystyle c(l)=1.51\;\frac{J}{g⋅K}\)
\(\displaystyle c(g)=1.70\;\frac{J}{g⋅K}\)
First, draw the heating curve. We see benzene is a liquid at 22°C and a gas at 90°C.
We need to determine the heat in regions CD, DE, and EF. Once we have calculated the heats, we then sum them.
\(\displaystyle Region_{CD}=m\times c_l \times ΔT\)
ΔT = 80.1°C – 5.5°C = 74.6°C = 74.6 K
\(\displaystyle Region_{CD}=25.0\;g\times 1.51 \frac{J}{g⋅K}\times 74.6\;K\;=\;2816\;J\;=\;2.816\;kJ\)
moles benzene = 25.0 g x (1 mole/78.11 g) = 0.320 moles benzene
\(\displaystyle Region_{DE}= ΔH_{vap}\times moles\;=\;33.2\frac{kJ}{mol}\times 0.320\;mol\;=\;10.6\;kJ\)
\(\displaystyle Region_{EF}= m\times c_g \times ΔT\)
ΔT = 90.0°C – 80.1°C = 9.9°C = 9.9 K
\(\displaystyle Region_{EF}=25.0\;g\times 1.70\frac{J}{g⋅K}\times 9.9\;K\;=\;420.75\;J\;=\;0.420\;kJ\)
ΔH = 2.816 kJ + 10.6 kJ + 0.420 kJ = 13.8 kJ
Back to Heating and Cooling Curves
Exercise 2. What is ΔS for ethanol if the boiling point is 73.37°C and ΔH = 42.3 kJ/mol?
73.37°C + 273.15 = 346.52 K
\(\displaystyle ΔS\;=\;\frac {ΔH}{T}=\frac {42.3\frac{kJ}{mol}}{346.52\;K}\;=\;0.122\frac{kJ}{mol⋅K}\;=\;\mathbf{122\frac{J}{mol⋅K}}\)
Back to Heating and Cooling Curves
Exercise 3. Answer the questions below about the following curve for a substance that is being heated.
A) At what temperature does the substance condense? 77°C
B) At what temperature does the substance freeze? -23°C
C) What phase(s) is present at Points A, B, D, and E?
At Point A there is solid present
At Point B there is both solid and liquid present
At Point D there is liquid and gas present
At Point E there is gas present
D) What equation is used to determine the heat in Region C?
q = mass x cl x ΔT
E) What equation is used to determine the heat in Region B?
q = moles x ΔHfus
F) What is the boiling point of the substance?
77°C
G) Why does the temperature stay constant in Region D?
The thermal energy is being used to disrupt the intermolecular forces of attractions
Back to Heating and Cooling Curves
Exercise 4. What is ΔH if acetone cools from 75.5°C to 25.0°C?
Melting Point = -95°C
ΔHfus = 5.70 kJ/mol
ΔHvap = 31.30 kJ/mol
cs = 1.653 J/(g⋅K)
cl = 2.409 J/(g⋅K)
cg = 1.283 J/(g⋅K)