1.10 Practice Problems

Attempt these problems as if they were real exam questions in an exam environment.

Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem.

Which of the following n–alkanes has the highest boiling point?
1. C₅H₁₂
2. C₇H₁₆
3. C₄H₁₀
4. C₉H₂₀
5. C₃H₈
Solution

Answer: D

Concept: Intermolecular forces and structure

The strength of intermolecular forces can affect the boiling points of a substance. Stronger IMFs generally lead to larger boiling points.

Given is a list of hydrocarbons which predominantly exhibit dispersion. Dispersion increases with molecule size. Therefore, the largest molecule in this series exhibts the most dispersion and, therefore, is expected to have the highest boiling point.
What word describes the phase transition from the liquid to the gas phase?
1. sublimation
2. fusion
3. condensation
4. vaporization
5. none of these
Solution

Answer: D

Concept: Phase transitions
Which of the following is considered to be an endothermic process?
1. fusion
2. condensation
3. deposition
4. freezing
5. none of these
Solution

Answer: A

Concept: Phase transitions

Fusion (also known as melting) requires an input of thermal energy (endothermic).
Predict the state of an unknown substance if the temperature was significantly decreased and the pressure was significantly increased.
1. solid
2. liquid
3. gas
Solution

Answer: A

Concept: Phase diagram

Generally, decreasing temperature of a substance results in a gas → liquid → solid.

Generally, increasing pressure of a substance results in a gas → liquid → solid.
The heat of vaporization for water at 100 °C is 40.7 kJ mol^–1. What is the heat of vaporization (in kJ mol^–1) for water at 25 °C?
1. exactly –40.7
2. greater than 40.7
3. equal to 40.7
4. less than 40.7
5. none of these
Solution

Answer: B

Concept: Heat/Enthalpy of vaporization

The heat of vaporization for a substance increases with decreasing temperature.
According to Hess’s Law, the enthalpy of sublimation can be approximated by adding which two terms together?
1. ΔH_frz + ΔH_dep
2. ΔH_frz + ΔH_vap
3. ΔH_sub + ΔH_dep
4. ΔH_fus + ΔH_vap
5. none of these
Solution

Answer: D

Concept: Phase transitions

A subtance heated from a solid to a gas sometimes melts (\(\Delta H_{\mathrm{fus}}\)) and then vaporizes (\(\Delta H_{\mathrm{vap}}\)). Sublimation can be approximated as the the addition of these two terms.

\[\Delta H_{\mathrm{sub}} \approx \Delta H_{\mathrm{fus}} + \Delta H_{\mathrm{vap}}\]
How much liquid water (in L) at 100 °C can be theoretically vaporized when 30,000 kJ worth of heat is absorbed? Assume water has a density of 1.0 g mL^–1. ΔH_vap(H₂O) = 43.46 kJ mol^–1
1. 12.44
2. 1.24 × 10⁴
3. 3.12 × 10^–3
4. 0.13
5. 4.6 × 10²
Solution

Answer: A

Concept: Heat/Enthalpy of vaporization

Enthalpy is an extensive property.

Moles of water that can be vaporized

\[\begin{align*} q &= n\Delta H_{\mathrm{vap}} \longrightarrow \\[1.5ex] n &= \dfrac{q}{\Delta H_{\mathrm{vap}}} \\[1.5ex] &= \dfrac{3.0\times 10^4~\mathrm{kJ}}{43.46~\mathrm{kJ~mol^{-1}}} \\[1.5ex] &= 690.289~\mathrm{mol~H_2O} \end{align*}\]

Moles to mass (in g)

\[\begin{align*} m &= n \times \text{molar mass} \\[1.5ex] &= 690.289~\mathrm{mol} \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 12,439.01~\mathrm{g} \end{align*}\]

Mass to volume (in L)

\[\begin{align*} d &= \dfrac{m}{V} \longrightarrow \\[1.5ex] V &= \dfrac{m}{d} \\[1.5ex] &= \left ( \dfrac{12,439.01~\mathrm{g}}{\mathrm{1.0~g~mL^{-1}}} \right ) \left ( \dfrac{1~\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[1.5ex] &= 12.44~\mathrm{L} \end{align*}\]
How much heat is evolved/released (in kJ) when 100 g of 80 °C water is converted into –35 °C ice? This process is carried out at sea level.
c_solid = 2.09 J g^–1 °C^–1, c_liquid = 4.184 J g^–1 °C^–1, c_gas = 1.84 J g^–1 °C^–1
ΔH_fus = 6.01 kJ mol^–1, ΔH_vap = 40.67 kJ mol^–1
1. 24
2. 33
3. 74
4. 105
5. 662
Solution
Answer: C

Concept: Heating curve

Recognize that the substance is water so we are expected to know its normal freezing and boiling points. If the substance was not water, these values would be given.

\[\begin{align*} \Delta T_{\mathrm{f}} &= 0~^{\circ}\mathrm{C} \\[1.5ex] \Delta T_{\mathrm{b}} &= 100~^{\circ}\mathrm{C} \end{align*}\]

Next, realize that three regions of the heating curve must be accounted for.
```
 1. Cooling liquid to freezing point
 2. Freezing the liquid
 3. Cooling the solid to the final temperature
```
Heat lost when cooling liquid to freezing point

\[\begin{align*} q_1 &= mc_{\mathrm{liquid}}\Delta T \\[1.5ex] &= (100~\mathrm{g})(4.184~\mathrm{J~g^{-1}~^{\circ}C})(0~^{\circ}\mathrm{C} - 80~^{\circ}\mathrm{C}) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}}\right )\\[1.5ex] &= -33.472~\mathrm{kJ} \end{align*}\]

Heat lost when undergoing freezing

Because water is freezing, ΔH_fus = –ΔH_frz. Also, convert mass to moles.

\[\begin{align*} q_2 &= n\Delta H_{\mathrm{frz}} \\[1.5ex] &= \left ( \dfrac{100~\mathrm{g}}{18.02~\mathrm{g~mol^{-1}}} \right ) \left ( -6.01~\mathrm{kJ~mol^{-1}} \right ) \\[1.5ex] &= -33.352~\mathrm{kJ} \end{align*}\]

Heat lost when cooling solid to final temperature

\[\begin{align*} q_3 &= mc_{\mathrm{liquid}}\Delta T \\[1.5ex] &= (100~\mathrm{g})(2.09~\mathrm{J~g^{-1}~^{\circ}C})(-35~^{\circ}\mathrm{C} - 0~^{\circ}\mathrm{C}) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}}\right )\\[1.5ex] &= -7.315~\mathrm{kJ} \end{align*}\]

Total heat lost

\[\begin{align*} q_{\mathrm{tot}} &= q_1 + q_2 + q_3 \\[1.5ex] &= -33.472~\mathrm{kJ} + -33.352~\mathrm{kJ} + -7.315~\mathrm{kJ} \\[1.5ex] &= -74.139~\mathrm{kJ} \end{align*}\]

Because the question is asking for “heat evolved/released”, rationalize this to be a positive value. (If the question asked for the change in heat of the water, we could leave this as a negative value).
The plot below gives the vapor pressures of benzene over a range of temperatures. The equation for the linear regression is also given. What is ΔH_vap (in kJ mol^–1) for benzene?

\[y = -4020.5x + 16.014\]
1. 15.83
2. 33.43
3. –8.67
4. 54.13
5. 28.67
Solution

Answer: B

Concept: Clausius-Clapeyron linear form

Recall that the ln(P) vs 1/T relation gives an approximately straight line whose slope, m, is defined by the heat of vaporization of a substance such that

\[m = -\dfrac{\Delta H_{\mathrm{vap}}}{R}\]

The slope of the line in the plot is in the given equation for the linear regression. The units of Kelvin are ignored here.

Find Heat of Vaporization

\[\begin{align*} m &= -\dfrac{\Delta H_{\mathrm{vap}}}{R} \longrightarrow \\[1.5ex] \Delta H &= -m \times R \\[1.5ex] &= -(-4020.5) \left( 8.314~\mathrm{J~mol^{-1}~K^{-1}} \right ) \left ( \dfrac{1~\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= 33.426~\mathrm{kJ~mol^{-1}} \end{align*}\]
A given solvent has a normal boiling point of 93.7 °C and ΔH_vap = 19.3 kJ mol^–1. What is the boiling point (in °C) of the solvent when the pressure is 947 mmHg (1.24605 atm)?
1. 81.4
2. 118.8
3. 102.4
4. 107
5. 94.5
Solution

Answer: D

Concept: Clausius-Clapeyron two-point form

We are provided with two pressures [normal (760 mmHg) and 947 mmHg] as well as one boiling point and a heat of vaporization (ΔH_vap). We are asked to solve for the second boiling point. We use the Clausius-Clapeyron two-point form equation.

Identify Variables

ΔH_vap = 19.3 kJ mol^–1
P₁ = 760 mmHg or 1 atm
P₂ = 947 mmHg or 1.24605 atm
T₁ = 93.7 °C = 366.85 K
T₂ = ? °C

Solve for T₂

\[\begin{align*} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) \longrightarrow \\[1.5ex] T_2 &= \dfrac{1}{\left (-\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_1}\right )}\\[1.5ex] &= \dfrac{1}{\left (-\dfrac{\ln \left ( \dfrac{1.24605~\mathrm{atm}}{1~\mathrm{atm}} \right ) } {\dfrac{19.3~\mathrm{kJ~mol^{-1}~K^{-1}}\left( \dfrac{10^3~\mathrm{J}}{1~\mathrm{kJ}}\right )}{8.314~\mathrm{J~mol^{-1}~K^{-1}}}} + \dfrac{1}{366.85~\mathrm{K}}\right ) } - 273.15~\mathrm{K}\\[1.5ex] &= 106.616~\mathrm{K} \end{align*}\]

Math String: 1/((-ln(1.24605/1)/(19.3*1000/8.134))+1/366.85)-273.15
Liquid benzene has a vapor pressure of 760 mm Hg and a heat of vaporization of 30.78 kJ mol^–1 at 80.0 °C. If the vapor pressure and temperature were plotted for benzene around this temperature using the linear form of the Clausius-Clapeyron equation, what would the slope of the resulting line be (to the nearest whole number)?
1. –3702
2. –4
3. –3.7
4. 1
5. 3700
Solution

Answer: A

Concept: Clausius-Clapeyron linear form

Recall that the ln(P) vs 1/T relation gives an approximately straight line whose slope, m, is defined by the heat of vaporization of a substance such that

\[m = -\dfrac{\Delta H_{\mathrm{vap}}}{R}\]

We must determine the slope using the given heat of vaporization.

Find the slope

\[\begin{align*} m &= -\dfrac{\Delta H_{\mathrm{vap}}}{R} \\[1.5ex] &= -\dfrac{30.78~\mathrm{kJ~mol^{-1}} \left ( \dfrac{10^3~\mathrm{J}}{1~\mathrm{kJ}} \right )} {8.314~\mathrm{J~mol^{-1}}} \\[1.5ex] &= -3702.189 \end{align*}\]
Toluene, a common organic solvent, has a heat of vaporization of 38.06 kJ mol^–1 and boils at 110.4 °C at sea level. What is the boiling point (in °C) of toluene at the top of Mount Fuji where the atmospheric pressure is only 65% as strong as the pressure at sea level?
1. 83
2. 88
3. 94
4. 97
5. 104
Solution

Answer: D

Concept: Clausius-Clapeyron two-point form

We are provided with two pressures [normal (1 atm) and 65% of 1 atm (0.65 atm)] as well as one boiling point and a heat of vaporization (ΔH_vap). We are asked to solve for the second boiling point. We use the Clausius-Clapeyron two-point form equation.

Identify Variables

ΔH_vap = 38.06 kJ mol^–1
P₁ = 1 atm
P₂ = 0.65 atm
T₁ = 110.4 °C = 383.55 K
T₂ = ? °C

Solve for T₂

\[\begin{align*} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) \longrightarrow \\[1.5ex] T_2 &= \dfrac{1}{\left (-\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{\Delta H_{\mathrm{vap}}}{R}} + \dfrac{1}{T_1}\right )}\\[1.5ex] &= \dfrac{1}{\left (-\dfrac{\ln \left ( \dfrac{0.65~\mathrm{atm}}{1~\mathrm{atm}} \right ) } {\dfrac{38.06~\mathrm{kJ~mol^{-1}~K^{-1}}\left( \dfrac{10^3~\mathrm{J}}{1~\mathrm{kJ}}\right )}{8.314~\mathrm{J~mol^{-1}~K^{-1}}}} + \dfrac{1}{383.55~\mathrm{K}}\right ) } - 273.15~\mathrm{K}\\[1.5ex] &= 97.318~\mathrm{K} \end{align*}\]

Math String: 1/((-ln(0.65/1)/(38.06*1000/8.134))+1/383.55)-273.15