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Heating Curves and Vapor PressureIntroduction – When studying Thermochemistry, you learned about relating the energy for

systems and surrounding and how to determine the amount of thermal energy exchange

between them. However, in those systems, you were not asked to consider the thermal energy

requirements for a system to change its physical state. In this activity you will explore the

thermal energy requirements for a substance to undergo any state (phase) change and will then

further explore the liquid-gas phase equilibrium as a function of intermolecular forces.

Prerequisite knowledge – This activity requires that you are comfortable with the following

skills and understanding. If you need to review this material, the page numbers provided

indicate where a topic is covered within the OpenStax (2e) text book.

Prerequisite knowledge

Quantifying thermal energy exchange

Reaction enthalpy

Relations involving change in enthalpy

Intermolecular forces

OpenStax

230-237

251-255

260-265

518-530

Changing temperature versus changing state – Most people are familiar with the temperatures

at which pure water changes states. They know that water melts/freezes at 0 C and

boils/condenses at 100 C. In studying Thermochemistry, you dealt with determining the

energy requirements to reach one of these state-change points (temperatures). However,

simply heating a sample of water to one of these temperatures does not instantly change its

state, additional energy is required to weaken or overcome the intermolecular forces between

water molecules – thereby allowing them to transition to the next state.

——————————————————————————————————————————Model 1 – Melting an ice cube

Thermal Energy

The graph below (called a heating curve) shows that as you add thermal energy to melt an ice

cube, the energy is used to both raise the temperature of the sample as well as to convert the

sample from the solid state to the liquid state.

Model 1 continues on next page.

Model 1 continued.

Energy requirements to convert an ice cube at -10.0 C to liquid water at +10.0 C

Energy required to raise

temperature of solid:

𝒒𝟏 = 𝒎𝑪𝒔,𝒔𝒐𝒍𝒊𝒅 ∆𝑻

Energy required to convert

from solid to liquid:

𝒒𝟐 = 𝒏∆𝑯𝒇𝒖𝒔𝒊𝒐𝒏

Energy required to raise

temperature of liquid:

𝒒𝟑 = 𝒎𝑪𝒔,𝒍𝒊𝒒𝒖𝒊𝒅 ∆𝑻

Note: you should be familiar with all of the variables and units in these equations from the last

chapter of material on Thermochemistry (Tro, chapter 6). The only new variable is Hfusion, which is the

enthalpy change for the melting (fusion) of a solid.

——————————————————————————————————————————Key Questions – Use the information provided in Model 1 to answer these questions.

KQ 1) According to the heating curve, at what temperature does the ice cube melt?

KQ 2) Did it require more or less thermal energy to raise the temperature of the ice cube to its

melting point or to convert it from a solid to liquid?

KQ 3) Write the equation used to determine the amount of thermal energy (q) required to raise

the temperature of a solid and define all of the variables (including their units) in the equation.

2

KQ 4) Write the equation used to determine the amount of thermal energy (q) required to

convert from a solid to a liquid and define all of the variables (including their units) in the

equation.

KQ 5) How is the equation for raising the temperature of a liquid different from that for raising

the temperature of a solid?

Exercises – These questions will help you develop your understanding of the relationships

presented in Model 1. For each Exercise, report numerical answers with the correct number of

significant digits.

EX 1) Determine the amount of thermal energy (q) required to raise the temperature of 10.0 g

𝐽

solid water from -10.0 C to its melting point. 𝐶𝑠,𝑠𝑜𝑙𝑖𝑑 = 2.09 𝑔∙℃

EX 2) Determine the amount of thermal energy (q) required to convert 10.0 g solid water to

𝑘𝐽

liquid water at its melting point. ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛 = 6.02 𝑚𝑜𝑙

EX 3) Determine the amount of thermal energy (q) required to raise the temperature of 10.0 g

𝐽

liquid water from its melting point to +10.0 C. 𝐶𝑠,𝑙𝑖𝑞𝑢𝑖𝑑 = 4.18 𝑔∙℃

EX 4) Determine the total amount of thermal energy required to raise 10.0 g of ice at -10.0 C to

liquid water at +10.0 C.

3

——————————————————————————————————————————Model 2 – Heating and cooling curve

By defining the freezing (Tf) and boiling (Tb) points, the generic curve shown below can be used

to investigate the heating (adding thermal energy) or cooling (removing thermal energy) for any

substance.

Note: the curve represents a general profile, sections are not to scale for any one substance.

The curve sections and their related energy equations for each are noted in the table.

Section

Adding thermal energy

Description

Equation

Section

Removing thermal energy

Description

Equation

1

Solid Phase

𝒒𝟏 = 𝒎𝑪𝒔,𝒔𝒐𝒍𝒊𝒅 ∆𝑻

5

Gas Phase

𝒒𝟓 = 𝒎𝑪𝒔,𝒈𝒂𝒔 ∆𝑻

2

Solid→Liquid

(melting)

𝒒𝟐 = 𝒏∆𝑯𝒇𝒖𝒔𝒊𝒐𝒏

*4

Gas→Liquid

(condensation)

𝒒𝟒 = −𝒏∆𝑯𝒗𝒂𝒑𝒐𝒓𝒊𝒛𝒂𝒕𝒊𝒐𝒏

3

Liquid Phase

𝒒𝟑 = 𝒎𝑪𝒔,𝒍𝒊𝒒𝒖𝒊𝒅 ∆𝑻

3

Liquid Phase

𝒒𝟑 = 𝒎𝑪𝒔,𝒍𝒊𝒒𝒖𝒊𝒅 ∆𝑻

4

Liquid→Gas

(vaporization)

𝒒𝟒 = 𝒏∆𝑯𝒗𝒂𝒑𝒐𝒓𝒊𝒛𝒂𝒕𝒊𝒐𝒏

*2

Liquid→Solid

(freezing)

𝒒𝟐 = −𝒏∆𝑯𝒇𝒖𝒔𝒊𝒐𝒏

5

Gas Phase

𝒒𝟓 = 𝒎𝑪𝒔,𝒈𝒂𝒔 ∆𝑻

1

Solid Phase

𝒒𝟏 = 𝒎𝑪𝒔,𝒔𝒐𝒍𝒊𝒅 ∆𝑻

*Note: The processes of vaporization and condensation as well as melting and freezing are

opposite processes. As H values are only reported for the endothermic processes, the

addition of a negative sign is required in using them for the corresponding exothermic

processes.

——————————————————————————————————————————4

Key Questions – Use the information provided in Model 2 to answer these questions.

KQ 6) The curve in Model 2 notes 4 points, labeled A-D. Describe the substance at each of these

points, point A has been done as an example.

Example: At point A, the substance is in its solid state and is at the freezing/melting transition

temperature.

KQ 7) Explain why moving from point C to point D on the curve does not require the inclusion

of a negative sign in the formula but moving from point D to point C does require a negative

sign in the formula.

KQ 8) Explain why there is no need to include a negative sign in the formula when moving from

point B to point C OR from point C to point B.

KQ 9) If a sample starts as a gas, with a temperature above its boiling point, how many

individual calculations of thermal energy (q) are required to determine the total amount of

thermal energy removed in converting the entire sample to a solid at its freezing temperature?

5

Exercises – These questions will help you develop your understanding of the relationships

presented in Model 2. For each Exercise, report numerical answers with the correct number of

significant digits. Use the generic curve in Model 2 and the tabulated data below when solving

these exercises.

Value (units)

𝑇𝑓 (℃)

𝑇𝑏 (℃)

𝑘𝐽

∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛 (

)

𝑚𝑜𝑙

𝑘𝐽

∆𝐻𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 (

)

𝑚𝑜𝑙

𝐽

𝐶𝑠,𝑠𝑜𝑙𝑖𝑑 (

)

𝑔∙℃

𝐽

𝐶𝑠,𝑙𝑖𝑞𝑢𝑖𝑑 (

)

𝑔∙℃

𝐽

𝐶𝑠,𝑔𝑎𝑠 (

)

𝑔∙℃

Water, H2O

0.00

100.0

Benzene, C6H6

5.40

80.1

6.02

9.90

40.7

30.7

2.09

1.51

4.18

1.73

2.01

1.33

EX 5) Determine the total amount of thermal energy that is evolved when a 10.0 g sample of

water vapor at 105.0 C is cooled to liquid water at 60.0 C.

EX 6) Determine the total amount of thermal energy required to convert a 25.0 g sample of

solid benzene at -2.40 C to benzene vapor at 108.0 C.

6

Vapor Pressure – In a sealed container at a given temperature, an

equilibrium is established between the vaporization and condensation

processes. The amount of gas above the liquid exerts a pressure called the

‘vapor pressure’ for the liquid.

The boiling point for a liquid is defined as the temperature at which the

liquid’s vapor pressure is equal to the external pressure (e.g., atmospheric

pressure). Hence, the boiling point of a liquid changes as the external

pressure changes. For example, pure water has a boiling point of 100.0 C

at an atmospheric pressure of 760 torr (‘sea level’). However, its boiling

point is elevation dependent and drops to 94.0 C in Denver, Colorado

(altitude = 5280 ft, approximate atmospheric pressure = 631 torr).

——————————————————————————————————————————Model 3 – Vapor pressure curves

The vapor pressure for four different pure liquids are plotted as a function of their

temperature. The curves vary due to the intermolecular forces present in each liquid.

——————————————————————————————————————————Key Questions – Use the information provided in Model 3 to answer these questions.

KQ 10) Which molecule has the weakest intermolecular forces present in its liquid sample?

KQ 11) Which molecule has the lowest normal boiling point?

7

KQ 12) If each of the pure liquids in Model 3 were heated to 20.0 C, which would have the

highest vapor pressure?

Exercises – These questions will help you develop your understanding of the relationships

presented in Model 3.

EX 7) Acetone (C3H6O) has a normal boiling point of 56.1 C. Compare the strength of its

intermolecular forces and its vapor pressure to those of ethyl alcohol.

EX 8) Use the information about three unknown compounds to complete each statement.

Compound A

Compound B

Compound C

Molar Mass (g/mol)

100.16

100.20

102.17

Normal Boiling Point (C)

128

98

159

i) Compound ____ experiences hydrogen bonding.

ii) Compound ____ is polar but unable to experience hydrogen bonding.

iii) Compound ____ is neither polar nor able to experience hydrogen bonding.

iv) Compound ____ will have the highest vapor pressure of the three compounds.

EX 9) Using Model 3, determine the normal boiling point of diethyl ether and ethyl alcohol

Normal Tboil

Tboil at 200 torr

diethyl ether

_______________°C

_______________°C

ethyl alcohol

_______________°C

_______________°C

——————————————————————————————————————————Model 3– More Information Vapor pressure vs. temperature curves

The relationship between vapor pressure and temperature is an exponential growth curve and

can be modeled by the Clausius-Clapeyron Equation. Manipulation of the Clausius-Clapeyron

Equation shows that the logarithm of the vapor pressure vs. inverse absolute temperature is a

linear function whose slope is proportional to the ΔHvap. Further manipulation can yield a form

of the Clausius-Clapeyron Equation in a two-point form (needing just two measurements of

8

vapor pressure and temperature), see Equation 1. The two-point form generally gives less

accurate results but is quite nice in a classroom setting. The two-point form can also be used to

estimate the boiling point at various ambient atmospheric pressures as long as you know ΔHvap

and a vapor pressure at a particular temperature (remember: the vapor pressure at the normal

boiling point is 760 torr).

P − H vap 1 1

−

ln 2 =

R T2 T1

P1

Equation 1: Clausius-Clapeyron Equation in a two-point form. P1, T1 and P2, T2 represent

some initial and final set of temperature and vapor pressure for a particular substance.

Pressure can be in any unit as long as P1 and P2 have the same units. Temperature must be

in units of Kelvin. ΔHvap is the enthalpy of vaporization in units of J/mol. R is the ideal gas

constant 8.314 J/mol-K. ln indicates that you will be taking the natural log of the value in

parentheses.

Ex 10) Use your values from EX 9 and equation 1 to estimate the ΔHvap for diethyl ether and

ethyl alcohol.

diethyl ether

ethyl alcohol

9

Challenge Problem – This problem requires you to combine your prior knowledge of

calorimetry with your newly acquired knowledge of phase change energy.

1. A 53.5 g sample of ice at exactly 0 C is placed in 115. g of water at 75. C. If no thermal

energy is lost to the surroundings, what is the final temperature of the mixture?

2. The ΔHvap for mercury is 59.1 kJ/mol. Its normal boiling point is 357°C. What is the

vapor pressure at 25°C? Express your answer to 2 sig figs.

3. At 25°C, how many grams of mercury vapor would fill up a room that is 3.00 m X 4.00

m x 5.00 m?

10