A sample of 1.435 g of naphthalene , a compound commonly used in mothballs, is completely burned in a constant-volume bomb calorimeter. During the combustion, the temperature of the calorimeter and the surrounding water increases from 20.28°C to 25.95°C. The combined heat capacity of the bomb calorimeter and water is 10.17 kJ °C⁻¹.

Calculate the molar heat of combustion of naphthalene.

Interpretation

A bomb calorimeter is designed to measure the heat released during combustion at constant volume. Since the calorimeter is well insulated, the heat released by the chemical reaction does not escape to the surroundings. Instead, it is absorbed entirely by the calorimeter and the water.

Thus, the heat lost by the reaction is exactly equal in magnitude to the heat gained by the calorimeter:
\( q_{{rxn}} + q_{{cal}} = 0 \)

or,

The negative sign indicates that combustion is an exothermic process, meaning the system releases heat while the calorimeter absorbs it.

Concept Application

The information provided allows us to calculate the heat absorbed by the calorimeter because both its heat capacity and the temperature rise are known.

The relationship is

where

• \( C_{{cal}} = \) heat capacity of the calorimeter,
• \( ΔT = \) Change in temperature

Once the heat released by 1.435 g of naphthalene is obtained, it can be converted to a per mole basis using the molar mass of naphthalene.

Solution

Step 1: Calculate the temperature change

The temperature change is

Step 2: Calculate the heat absorbed by the calorimeter

Using \( q_{{cal}} = C_{{cal}}× ΔT \)

we obtain

This is the amount of heat absorbed by the calorimeter.

Since

the combustion of 1.435 g of naphthalene releases

Step 3: Calculate the molar mass of naphthalene

The molar mass is

Step 4: Convert the heat released to one mole

Using the conversion

we get

Hence, the molar heat of combustion of naphthalene is \( -5.151 \times 10^3 kJ mol^-1\)

The negative sign signifies that heat is released during combustion.

Insight

A useful idea to remember is that in calorimetry, the heat released by the reaction is always equal in magnitude and opposite in sign to the heat absorbed by the calorimeter:

Also, notice that calorimeters measure the heat released by a known mass of substance. To report the molar heat of combustion, always convert the given mass into one mole using the molar mass. This two-step approach ,first finding the heat for the given sample, then converting it to one mole works for nearly every combustion calorimetry problem.

Calculate the change in internal energy, \(ΔE\), when 2 moles of carbon monoxide are converted into 2 moles of carbon dioxide at 1 atm and $25^\circ\text{C}$

Given:

\( ΔH=−566.0 kJ/mol\)
$$$$

Interpretation

This problem asks us to determine the change in internal energy \(ΔE\) from the given enthalpy change \(ΔH\).

Although both quantities measure the energy change during a chemical reaction, they are generally not equal for reactions involving gases. The reason is that gases can expand or contract, causing the system to perform pressure–volume (PV) work.

The relationship between enthalpy and internal energy at constant pressure is

or, equivalently,

where

• \(Δn_g​\) = change in the number of moles of gaseous species
• \(R\) = universal gas constant
• \(T\) = temperature in kelvin

The correction term \(Δn_g​RT\) accounts for the energy associated with the expansion or contraction of gases.

Concept Application

The first step is to calculate the change in the number of gaseous moles.

The balanced reaction is \(2CO(g) + O_2(g) \rightarrow 2CO_2 (g)\)

The reactant side contains

while the product side contains \(2  moles   of  gas\)

Hence,

Notice that only gaseous species are counted. Solids and liquids are ignored because they do not contribute significantly to expansion work.

Since the given enthalpy is expressed in kilojoules, the correction term must also be expressed in kilojoules. Therefore, after calculating \(RT\) in joules, we convert it to kilojoules.

Solution

Using the equation

Substituting the known values:

First calculate the correction term:

Converting joules into kilojoules,

Now substitute into the equation:

Rounding to one decimal place,

Final Answer

$$\boxed{{\displaystyle }}$$\( ΔE =-563.5 kJ/mol\)

​

Insight

Whenever a thermochemistry problem asks you to convert between enthalpy and internal energy, first check whether the number of gaseous moles changes.

• If \(Δn_g​<0\), the number of gas molecules decreases, so \( ΔE\)is less negative than \( ΔH\)
• If \(Δn_g​>0\), the number of gas molecules increases, so the number of gas molecules decreases, so \( ΔE\)is more negative than \( ΔH\)

Exam Tip: Before substituting into the formula, always count only gaseous reactants and products. Solids and liquids are never included in the calculation of \(Δn_g​\), which helps avoid one of the most common mistakes in thermochemistry.

A gas is compressed inside a cylinder, and 462 J of work is done on the gas. During the same process, the gas releases 128 J of heat to the surroundings.

Using the first law of thermodynamics, calculate the change in the internal energy of the gas.

Interpretation

This problem is based on the First Law of Thermodynamics, which relates heat, work, and the change in the internal energy of a system.

The governing equation is:

where:

• \(\Delta E\) = change in internal energy of the system
• \(q\) = heat exchanged with the surroundings
• \(w\) = work done on or by the system

The most important part of such questions is assigning the correct signs to \(q\) and \(w\).

• If the system absorbs heat, then \(q > 0\).
• If the system releases heat, then \(q < 0\).
• If work is done on the system (compression), then \(w > 0\).
• If the system does work on the surroundings (expansion), then \(w < 0\).

A common mistake is to ignore these sign conventions. The calculation itself is simple once the signs are chosen correctly.

Concept Application

Here, the system is the gas.

The gas is compressed, meaning the surroundings push on the gas. Since work is done on the system,

The gas also releases heat to the surroundings. Heat leaving the system means

Now both quantities are ready to substitute into the First Law.

Solution

Using the First Law of Thermodynamics,

Substitute the given values:

Therefore,

The internal energy of the gas increases by \(+334\ \text{J}\).

Insight

A simple way to remember the sign convention is:

• Compression → Work done on the system → Positive \(w\)
• Expansion → Work done by the system → Negative \(w\)
• Heat enters → Positive \(q\)
• Heat leaves → Negative \(q\)

Think of internal energy like the balance in a bank account. Heat entering and work done on the system are deposits, while heat leaving and work done by the system are withdrawals. The net balance gives \(\Delta E\).

A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands:

(a) against a vacuum

(b) against a constant pressure of 1.2 atm

Interpretation

This problem tests the concept of pressure–volume work done during the expansion of a gas.

When a gas expands, it pushes against its surroundings. If the surroundings exert an external pressure P_ext , the gas must do work to move them. The work done is given by$$\boxed{w=-P_{\text{ext}}\Delta V}$$where

• \(w\) = work done by the system (gas)
• \(Pext​\) = external pressure
• \(ΔV=Vf​−Vi​\) = change in volume

The negative sign is very important. It follows the chemistry sign convention:

• Expansion: the gas does work on the surroundings, so \(w<0\).
• Compression: the surroundings do work on the gas, so \(w>0\).

A common misconception is to use the pressure of the gas in the formula. The work depends on the external opposing pressure, because that is the force the gas must overcome.

Concept Application

The gas expands from$$V_i=2.0\ \text{L}$$

to$$V_f=6.0\ \text{L}$$

Therefore,$$\Delta V=6.0-2.0=4.0\ \text{L}$$

The temperature is constant, but notice that no gas law calculation is needed. Since the external pressure is already given, the work can be calculated directly from the pressure–volume equation.

Solution

(a) Expansion against a vacuum

In a vacuum,$$P_{\text{ext}}=0$$

Substituting into the work equation,

$$w=-P_{\text{ext}}\Delta V$$$$w=-(0)(4.0\ \text{L})$$$$\boxed{w=0}$$

Reasoning

Although the gas expands, there is nothing pushing back against it. Since there is no opposing force, the gas performs no mechanical work on the surroundings.

(b) Expansion against a constant pressure of 1.2 atm

Here,$$P_{\text{ext}}=1.2\ \text{atm}$$

Using the work equation,$$w=-P_{\text{ext}}\Delta V$$$$w=-(1.2\ \text{atm})(4.0\ \text{L})$$$$w=-4.8\ \text{L}\cdot\text{atm}$$

The SI unit of work is joule, so convert using$$1\ \text{L}\cdot\text{atm}=101.3\ \text{J}$$

Therefore,$$w=-4.8\times101.3$$$$w=-486.2\ \text{J}$$

With appropriate significant figures,$$\boxed{w=-4.9\times10^2\ \text{J}}$$

Final Answers

(a) Against a vacuum$$\boxed{w=0}$$

(b) Against a constant pressure of 1.2 atm$$\boxed{w=-4.9\times10^2\ \text{J}}$$

Insight

A simple way to remember pressure–volume work is:

• No external pressure ⇒ no work, regardless of how much the gas expands.
• The work depends on the external pressure, not the gas pressure.
• Keep the sign convention in mind:
  • Expansion $\rightarrow$ gas loses energy by doing work → w<0.
  • Compression $\rightarrow$→ surroundings do work on the gas $\rightarrow$→ w>0.

A useful memory aid is:

$$\boxed{\text{Expansion} \;\Longrightarrow\; \text{Energy leaves the system as work} \;\Longrightarrow\; w<0}$$

This sign convention appears repeatedly in thermodynamics, so mastering it here will make later topics much easier.

A Bunsen burner flame takes 8 minutes to raise 250 g of water from 20.0°C to 100.0°C when the water is contained in a beaker of mass 100.0 g having a specific heat capacity of 0.670 J g⁻¹ °C⁻¹.

How long will it take to raise the temperature of 200.0 g of glycerine (specific heat capacity = 2.43 J g⁻¹ °C⁻¹) contained in the same beaker from 20.0°C to 80.0°C?

Interpretation

This problem is really about heat energy and heating rate.

A Bunsen burner supplies heat at a constant rate. That means:$$\text{Rate of heat supplied}=\frac{\text{Heat energy delivered}}{\text{Time}}$$

Since the same burner is used in both experiments, its heating rate remains unchanged. Therefore, if we know how much heat the burner delivers per minute in the first case, we can determine how long it will take to supply the heat required in the second case.

A common mistake is to calculate the heat needed only for the liquid. The problem clearly states that the glycerine is contained in the same beaker, so the beaker must also be heated. The heat absorbed by the beaker must therefore be included in both situations.

Concept Application

The heat required to raise the temperature of a substance is

$$q = mc\Delta T$$

where

• $m$ = mass
• $c$ = specific heat capacity
• $\Delta T$ = temperature change

The burner’s power can be found from the first experiment:

1. Calculate the total heat absorbed by the water and beaker.
2. Divide by 8 minutes to obtain the heat supplied per minute.
3. Calculate the total heat required for the glycerine and the same beaker.
4. Divide by the burner’s heating rate to obtain the required time.

Solution

Step 1: Heat absorbed by water

For the water,$$m_w = 250\,\text{g}$$$$\Delta T_w = 100-20 = 80^\circ\text{C}$$

Hence,$$q_w=(250)(4.18)(80)$$ $$q_w=83600\ \text{J}$$

Step 2: Heat absorbed by the beaker in the first experiment

For the beaker,$$m_b = 100.0\,\text{g}$$$$\Delta T_b = 80^\circ\text{C}$$

Therefore,$$q_b=(100.0)(0.670)(80)$$ $$q_b=5360\ \text{J}$$

Step 3: Total heat supplied in 8 minutes

$$q_{\text{total,1}} = q_w+q_b$$ $$=83600+5360$$$$=88960\ \text{J}$$

Thus the burner supplies$$\text{Heating rate} = \frac{88960}{8}$$$$=11120\ \text{J min}^{-1}$$

Step 4: Heat required for glycerine

For glycerine,$$m_g=200.0\,\text{g}$$$$\Delta T_g=80-20=60^\circ\text{C}$$

Hence,$$q_g=(200.0)(2.43)(60)$$$$q_g=29160\ \text{J}$$

Step 5: Heat required for the beaker in the second experiment

The beaker again rises from $20^\circ\text{C}$ to $80^\circ\text{C}$, so

$$\Delta T_b=60^\circ\text{C}$$$$q_b=(100.0)(0.670)(60)$$$$q_b=4020\ \text{J}$$

Step 6: Total heat needed

$$q_{\text{total,2}} = 29160+4020$$$$=33180\ \text{J}$$

Step 7: Calculate the time

$$t = \frac{q_{\text{total,2}}} {\text{Heating rate}}$$$$= \frac{33180} {11120}$$$$=2.98\ \text{min}$$$$\boxed{t \approx 3.0\ \text{min}}$$

Insight

The key idea is that time is proportional to the total heat required when the heating source remains the same:$$t \propto q$$

Also remember that whenever a container’s mass and specific heat are given, it usually means the examiner expects you to include the container’s heat absorption. Ignoring the beaker often gives an answer that is close, but not fully correct.$$\boxed{\text{Time required} \approx 3\ \text{minutes}}$$

Determine whether the entropy change is positive or negative for each of the following reactions, and explain the reasoning behind your predictions.

1. \(KClO_4(s) \rightarrow 2KClO_3(s)+O_2 (g)\)
2. \(H_2O(g) \rightarrow H_2O (l)\)
3. \(2Na(s) + 2H_2O(l)\rightarrow 2NaOH (aq)+H_2(g)\)
4. \(N_2(g) \rightarrow 2N (g)\)

Interpretation

Entropy $(S)$ is a measure of the number of possible microscopic arrangements (or the degree of dispersal of matter and energy) in a system. A process that produces greater molecular freedom generally increases entropy, whereas a process that creates a more ordered arrangement decreases entropy.

A very useful guideline is:$$S_{\text{solid}} < S_{\text{liquid}} < S_{\text{gas}}$$

because particles in gases have far greater freedom of movement than those in liquids or solids.

For many reactions, the change in the number of gaseous particles provides a quick clue:

$$\Delta n_{\text{gas}} = (\text{moles of gaseous products}) – (\text{moles of gaseous reactants})$$

An increase in gaseous particles usually means a positive entropy change, while a decrease usually means a negative entropy change. However, we should always interpret the physical change as well, not rely blindly on the formula.

(1) \(KClO_4(s) \rightarrow 2KClO_3(s)+O_2 (g)\)

Concept Application

The reactants contain only solids, while the products contain solids and a gas.

The formation of gaseous oxygen introduces particles with much greater freedom of motion than the solid reactants had.

Solution

$$\Delta n_{\text{gas}} = 1-0 = +1$$

Since gaseous molecules are produced, disorder increases significantly.$$\boxed{\Delta S > 0}$$​

Entropy change is positive.

(2) \(H_2O(g) \rightarrow H_2O (l)\)

Concept Application

This is a condensation process in which water vapor becomes liquid water.

Gas molecules move freely throughout a container, whereas liquid molecules are much more restricted and remain close together.

Solution

$$\Delta n_{\text{gas}} = 0-1 = -1$$

The transition from gas to liquid reduces molecular freedom and randomness.$$\boxed{\Delta S < 0}$$

Entropy change is negative.

(3) \(2Na(s) + 2H_2O(l)\rightarrow 2NaOH (aq)+H_2(g)\)

Concept Application

The reactants contain a solid and a liquid, but the products include hydrogen gas.

The appearance of a gaseous product greatly increases disorder because gas particles occupy a much larger volume and have many more possible arrangements.

Solution

$$\Delta n_{\text{gas}} = 1-0 = +1$$

Since gas is formed during the reaction, entropy increases.$$\boxed{\Delta S > 0}$$

Entropy change is positive.

(4) \(N_2(g) \rightarrow 2N (g)\)

Concept Application

Here, one gaseous molecule breaks into two gaseous atoms.

Even though both sides are gases, the number of independent particles doubles. More particles mean more possible arrangements and therefore greater disorder.

Solution

$$\Delta n_{\text{gas}} = 2-1 = +1$$

Because the number of gaseous particles increases,$$\boxed{\Delta S > 0}$$

Entropy change is positive.

Insight

A powerful way to predict entropy changes quickly is to ask two questions:

1. Is a gas being produced or consumed?
   Producing gas usually gives $\Delta S > 0$; consuming gas usually gives $\Delta S < 0$.
2. Has the number of gaseous particles increased?
   More gaseous particles mean more possible arrangements and therefore higher entropy.

In fact, reactions (a), (c), and (d) all increase entropy for the same underlying reason: the system ends up with more freedom of motion in the gaseous state. Reaction (b) is the opposite case, where that freedom is lost during condensation.