10.E: Thermochemistry (Exercises)

Q10.34a

Observe the following exchanges of energy. Identify whether each exchange is an example of heat or work. For each system, determine whether ΔE is positive or negative.

1. A cake is baked in an oven (The cake is the system)
2. Wood is burned for a fire (The wood is the system)
3. A boy kicks a chair, it is displaced in the direction of force applied. (The chair is the system)

Solution

1. The system is an example of heat, and because the cake gains heat, ΔE is positive.
2. The system is an example of heat, and because wood gives off heat, ΔE is negative.
3. The system is an example of work, and because work is done on the chair, ΔE is positive.

Q10.34b

Identify each energy exchange as primarily heat or work and determine the sign of \(\Delta E\) (positive or negative) for the system.

1. A soccer ball strikes into another soccer ball causing the soccer ball to stop rolling (The first soccer ball is defined as the system).
2. A military aircraft drops a bomb from the airplane, as the bomb drops, it hits the ground. (The bomb is defined as a system).
3. Two little girls is on a seesaw (the seesaw and the girls are defined as the system).

Definitions

• Heat : The transfer between two objects due to the temperature difference
• Work : A force acting over distance
• Energy: The ability of a system to do work or produce heat
• Kinetic Energy: when an object gains extra energy due to its motion
• Potential Energy: Can be converted to different forms of energy and still be able to do work in the process

Equations

• Relationship between Internal Energy \(\Delta E\), Heat (\(q\)), and Work (\(w\))

\[\Delta E = q+w\]

• Energy Flow between System and Surroundings

\[\Delta E_{sys}=-\Delta_{surr}\]

Form of Energy for Heat

• Heat-----> Kinetic Energy
• Kinetic Energy ----> Potential Energy
• Potential Energy---> Kinetic Energy

Strategy

1. Determine whether if each question is performing heat or work
2. Once you figured out if each system is producing heat or work, you can then determine whether your system is positive or negative
3. You're all done

Solutions

A.) Work,

So since the first soccer ball collided into the second soccer ball, we can define that it is an example of work because the first soccer ball had to have had some type of force for it to be able to collide into the second soccer ball.

Now that we have concluded that the soccer ball is producing work, we can also conclude that the delta E is negative because before the soccer ball was kicked, it started off as Kinetic Energy. But once it struck the second soccer ball, it released all of that energy it had formed and created potential energy .

B.) Work,

The reason why this would be considered to be work is because when the bomb was getting ready to be dropped it was able to show some type of action before it hit the ground.

The reason why the Delta E would be considered negative is because before dropping of the bomb, the bomb was considered as potential energy due to the energy being stored in the object. But once it was dropping, it formed kinetic energy and was able to gain extra energy due to the motion of the bomb dropping from the airplane.

C Work,

The reason why this would be considered as work is because as the little girls are going up and down on the seesaw, they are able to use a force to motion the seesaw.

The reason why this is considered to be positive is because both of the little girls are losing energy due to each one using force.

Example: Little Girl A ------releases energy to Little girl B so she can be pushed up on the seesaw causing Little Girl A to be on the ground. Then Little Girl B----releases all of her energy that was given to her and is able to transfer that energy back to Little Girl A causing her to be in the air while Little Girl B is back on the ground.

Q10.38

The air inside of a blimp is heated to create lift. The blimp as a system expands to do 81 kJ of work, and absorbs 150 J of heat,. What is the change of the internal energy of the system? (Answer to be provided in kJ)

Strategy

1. Identify the variables that we are given, which are: Heat (q) and Work (w). Where q=150 (J) and w=81 (kJ)
2. Identify what we are searching for, which is: Change in Internal Energy \[\Delta E\].
3. Realize that the equation required for this problem is \[\Delta E=q+w\]
4. The problem requests for an answer in (kJ), so we need to convert Heat (q) from (J) to (kJ). This is done through dimensional analysis \[(\dfrac{150J}{1})*(\dfrac{1kJ}{1000J})=.15kJ\] So, we have 0.15 (kJ) of Heat (q).
5. Now that we have our units correct we must check to see whether or not the Work (w) done is negative or positive. Work (w) is positive when work is done ON the system, however Work (w) is negative when work is done BY the system. The question defines the system as the blimp, which means that the work must be negative since the work is done BY the system as the blimp expands. So Work (w) is -81 kJ.
6. Now that we have correct units and determined the work to be negative we can plug the variables into the equation. \[\Delta E=(0.15kJ)+(-81kJ)\] So, \[\Delta E=-80.85kJ\] Thus, the change in internal energy is -80.85kJ

Answer

-80.85 (kJ)

Q10.44

The initial temperature for each substance, with an unknown mass, is 32 °C and absorbs 2450 J of heat. Below, each final temperature is recorded with each substance. With what is given, what is the mass of each substance?

1. Silver at 81.2 °C
2. Water at 54.3 °C
3. Granite at 68.9 °C

What we know

• Initial Temperature()
• Heat absorbed (2450 J)
• Final Temperatures (Silver:, Water:, Granite:)
• Specific Heats() (Silver: , Water:, Granite:).

What it is asking for: The mass of each substance

Strategy:

1. Find an equation suitable with the information given
2. Plug in all the information given in the appropriate places in the equation
3. Solve and get the mass for each substance

Solution

Heat= m

m=mass = specific heat =()

A. Silver

B. Water

C. Granite

Q10.41

What amount of energy needs to be removed from 0.75 pounds of silver to cool it from 110°F to 75°F? Assume silver has a density of 10.49 \[\dfrac{g}{cm^{3}}\]

Strategy

Use equation 10.5 on pg. 351 \[q= m\cdot C_{s}\cdot \Delta T\]

From Table 10.2 on pg. 351, for silver C_s = 0.235 \[\dfrac{J}{g\cdot ^{\circ}C}\]

First convert 0.75 pounds of silver to grams.

\[0.75 lb\cdot 453.6\dfrac{g}{lb}=340.2 g\]

Convert temperatures from °F to °C with the conversion °C = (X°F – 32°F)/1.8

110°F – 32°F/1.8 = 43.3°C

75°F – 32°F/1.8 = 23.9°C

Using equation 10.5 \[q=340.2g\cdot 0.235\dfrac{J}{g\cdot C} \cdot \left ( 43.3 C-23.9 C \right )=1551 J\]

Q10.43

20 g of each substance is initially at \[25^{\circ}C\]. If 2.10kJ of heat is absorbed, what is the final temperature for each substance?

1. Lead
2. Copper
3. Iron
4. Ethanol

Strategy

1. First, convert the amount of heat absorbed from 2.10 kJ to J by using the conversion factor 1 kJ/1000 J.
2. Next, find the specific heat capacities of each substance. This will be used in the formula.
3. Finally, use the formula “heat = mass X specific heat capacity X change in temperature” and plug in all the information known leaving the change in temperature as (x-25).
4. Solve for x.

Solution

1. \[2.10kJ \cdot \dfrac{1000J}{1kJ}= 2100J\]
2. 1. \[Lead= 0.128 \dfrac{J}{g} \cdot ^{\circ}C\]
   2. \[Copper= 0.385 \dfrac{J}{g} \cdot ^{\circ}C\]
   3. \[Iron= 0.449 \dfrac{J}{g} \cdot ^{\circ}C\]
   4. \[Ethanol= 2.42 \dfrac{J}{g} \cdot ^{\circ}C\]
3. 1. Lead \[2100J=20g \cdot 0.128\dfrac{J}{g \cdot^{\circ}C} \cdot (x-25)\]
   2. Copper- \[2100J=20g \cdot 0.385\dfrac{J}{g \cdot^{\circ}C} \cdot (x-25)\]
   3. Iron- \[2100J=20g \cdot 0.449\dfrac{J}{g \cdot^{\circ}C} \cdot (x-25)\]
   4. Ethanol- \[2100J=20g \cdot 2.42\dfrac{J}{g \cdot^{\circ}C} \cdot (x-25)\]
4. 1. Lead- Final temperature= \[845^{\circ}C\]
   2. Copper- Final temperature= \[298^{\circ}C\]
   3. Iron- Final temperature= \[258^{\circ}C\]
   4. Ethanol- Final temperature= \[68^{\circ}C\]

Q10.49

Calculate ΔE and ΔH for the evaporation of 1 mol of H₂O at a constant pressure. The reaction exerts 2468 kJ of heat and does 14 kJ of work.

1. ΔE
2. ΔH

Strategy

1. Realize ΔE_system= q + w (where q= heat and w=work), and the system is the material or process within which we are studying the energy changes in and the surrounds are everything else with which the system can exchange energy with. ΔE is a measure of the system's total energy.
2. Realize ΔE_universe = ΔE_system + ΔE_surrounding and ΔE_universe is constant, therefore when added together ΔE_system and ΔE_surroundings should add up to 0.
3. When energy flows out of a system ΔE_system <0, and when energy flows into a system ΔE_system>0
4. Realize at a constant pressure ΔH = q_p where q_p is heat at a constant pressure. ΔH is defined as the summation of the system's internal energy and the product of its pressure and volume. Since the pressure in this case is constant it is equal to the negative work and cancels out with the work from the energy equation giving us the equation ΔH = q_p. q is equal to the heat.

A. Realize that 2468 kJ is the amount of heat the reaction gives off and 14 kJ is the amount of work the reaction does

B. Realize that the reaction is both giving off heat and exerting work therefore the values will be negative

C. Take the values of both heat and work and add them together to get ΔE.

ΔE_system= (-2468 kJ) + (-14 kJ)

ΔE_system= -2482 kJ of work

D. Since ΔH = q_p, then q_p= -2468

Q10.50

Ethane has a change in enthalpy of 5500.2 kJ when 1.0 mol is combusted at 1.0 atm with a change of internal energy of 5512.4 kJ. Calculate the amount of work done during combustion.

Strategy and hints:

1. write out down the "givens" in the problem
2. choose the correct equation to determine the answer with the information you are given
3. make sure your signs are correct

Solution

ΔE_system=q+w ΔH = ΔE +ΔPV w= -PΔV

so, ΔE=ΔH+w

w=ΔH-ΔE

ΔE= 5512.4 kJ

ΔH= 5500.2 kJ

w= 5500.2 kJ - 5512.4kJ = -12.2 kJ

Helpful resources

• http://chemwiki.ucdavis.edu/Physical...s_in_Reactions
• http://chemwiki.ucdavis.edu/Physical...y_Calculations

Q10.52

Use the sign ∆H to indicate whether the following processes are endothermic or exothermic.

1. formation of ice
2. the hand touching hot stove
3. a lit candle flame

Strategy

1. Identify the system
2. Determine where the energy is moving
   • exothermic: moving out of the system into surroundings
   • endothermic: moving into the system into surroundings
3. Depending on whether the reaction is exothermic or endothermic, determine whether the reaction has a -∆H or +∆H.

Solutions

Solution a-formation of ice:

1. system: ice/water. Why? The object in the statement that was losing or gaining energy was the ice. Before the gain/lose of energy, the ice was water.
2. The ice is forming, which means that the water it once was is getting colder as water must reach 0˚C to begin to form ice. So, the water must lose this energy to become colder. Therefore, energy is leaving the system in order to form this ice. Because energy is leaving the system, the reaction is exothermic.
3. The reaction is exothermic, so the system is losing energy. Therefore, the ∆H of the system is going to be negative

Solution b-the hand on a hot stove:

1. system: hand. Why? The objects in the statement that was losing or gaining energy were the hand and the stove. However, the wording indicates that the object is the hand being affected by the stove, instead of the stove itself. This makes the hand the system and the stove its surroundings.
2. When a hand is placed onto a hot stove, it becomes very hot and may cause a burn to the skin. The hand is gaining that heat (AKA energy). Therefore, energy is entering the system, and therefore, it is endothermic.
3. The reaction is endothermic, so the system is gaining energy. Therefore, the ∆H of the system is going to be positive

Solution c-lit candle flame

1. system: egg. Why? In the statement, the only object that was stated gaining/losing energy was the flame
2. A fire is typically used to heat things up since it releases heat, which is a form of energy. This candle flame is emitting this energy from the system into the surrounding environment. Therefore, this process is exothermic.
3. Because it is exothermic, the system is losing its original amount of energy, which makes the ∆H negative.

Q10.53

Consider the equation for the combustion of methane (CH₄), a natural gas.

CH_4(g) + 2O_2(g)→ CO_2(g) + 2H₂O Delta H_rxn = -807.3kJ

If a bottle contains 200 mL of methane, how much heat is released by its complete combustion? The density of methane is 0.716 g/mL.

Strategy

1. Set up a conversion factor from the given volume of 200 mL and density of 0.716 g/mL to determine the mass of methane.
2. Set up a conversion factor to determine moles of methane from the mass.
3. Set up a conversion factor to determine the kJ of heat released from the ratio of moles of methane to (\Delta_{rxn}\).

Solution

A) 200mL CH₄ X 0.716 g/ML CH₄ 1 mol CH₄ X 807.3kJ CH₄

1 mL CH₄ 16.04g CH₄ 1 mol CH₄ = 7207.3 kJ

Q10.55

How much heat is generated after a complete combustion of of 6.50 kg of methane (\(CH_{(4(g)}\))?

\[CH_{(4(g)} + 2O_{2(g)} \rightarrow CO_{(2(g)} + 2H_2O_{(g)}\]

The molar enthalpy change for this reaction is \(\Delta_{rxn} = -1,250 \; kJ/mol\)

Solution

Step 1: Is the equation balanced? Yes. Also note that there is 1 mol of in this reaction. There are 2 mol of in this reaction. There is 1 mol of in this reaction. There are 2 mol of in this reaction.

Step 2: What is given? We know that there is 6.50 kg of .

We know that

Step 3: What other conversion factors might be useful to solve this problem?

1,000g=1kg

The molar mass of =16.05g/mol (Use the periodic table to locate the molar mass for and . Multiply the molar mass of hydrogen by 4. Add this number, 4.04g/mol, to the molar mass of ,12.01g/mol.)

Step 4: Set up dimensional analysis, starting with the value given in the question. (Make sure that your units "match" when completing this step.

The negative sign indicates that this is an exothermic reaction. In other words, the system (the reactants) lost energy to the surroundings (the products) in the form of heat.

10.60

As a lab experiment the class submerged a 47.3-g aluminum block with an initial temperature of 32.4 °C into an unknown mass of water at 73.2 °C. The temperature of the final mixture at equilibrium is 57.3 °C. Find the unknown mass of the water used in the experiment.

Definitions

Specific Heat (C_s): the required amount of heat energy to increase the temperature of one gram of a certain substance by 1 °C.

Helpful Equations:

Heat gained by aluminum = heat lost by water

\[Heat(Q)=mass\times C_{s}\times \Delta T\]

*Hint: Do not convert temperature to Kelvin

*Hint: Change in Temperature = T_final - T_initial

Specific heat of water = 4.18 J/g x C

Specific heat of aluminum = 0.903 J/g x C

Units:

Mass = grams

Specific Heat = J/g x C

Temperature = Celsius

Strategy

1. Due to the Law of Conservation of Energy, the heat exchange of aluminum and water can be set equal to each other. We know that aluminum will gain heat while water looses heat until they reach an equilibrium temperature.
2. Set up the equation as: \[mass_{water}\times C_{swater}\times \Delta T_{water}=-mass_{Al}\times C_{sAl}\times \Delta T_{Al}\]
3. Now plug in known values and solve.

Solution

\[mass_{water}\times 4.18J/g\cdot ^{\circ}C\times (53.7^{\circ}C-73.2^{\circ}C)=-47.3g\times (0.903J/g\cdot ^{\circ}C)\times (53.7^{\circ}C-32.4^{\circ}C)\]

\[Mass_{water}=\dfrac{-47.3g\times 0.903J/g\cdot ^{\circ}C\times (53.7^{\circ}C-32.4^{\circ}C)}{4.18J/g\cdot ^{\circ}C\times (53.7^{\circ}C-73.2^{\circ}C)}\]

\[mass_{water}=11.16g\]

Q10.66

In order to obtain the smallest possible amount of heat in which there is a large decrease in the number of moles of gas, should you carry out the reaction under conditions of constant volume or constant pressure? Explain.

Solution

In order to obtain the smallest amount of heat in which there is a large decrease in the number of moles of gas, you should carry out the reaction under conditions of constant volume. According to Avogadro’s Law (V₁/n₁=V₂/n₂), volume is directly proportional to the number of gas molecules. Thus, as the volume of the gas increases, the number of moles increases. Likewise, as the volume of the gas decreases, the number of moles decreases. As a result, if you wanted to obtain the smallest amount of heat in which the number of moles decreases, you should carry out the reaction with a decreasing constant volume.

Q10.67a

13.7 g of oxygen gas (O2) combusts in a bomb calorimeter. While undergoing combustion, the temperature rises from 44.1 °C to 48.4 °C. Computed in a completely different experiment, the heat capacity of the bomb calorimeter was found to be 6.24 kJ/°C. Calculate the ΔE for the burning of 1 mol of oxygen gas in kJ/mol oxygen.

1) ΔE_rxn=

Strategy

1. Bomb calorimeters are used to measure ΔE, or since it is a constant volume system. Since the combustion is under constant pressure the equation of ΔE_surroundings = q + w, where work is equal to negative the product of pressure and change in volume (w = -PΔV), is as follows ΔE_surr = q_cal
2. To calculate the ΔT or change in temperature, subtract the initial temperature from the final temperature.
3. Realize that heat (q) in the bomb calorimeter is equal to the negative heat of the system, but since it's not possible to measure the temperature of the system, we measure the change in temperature of the surroundings. q_cal = C_cal • ΔT.
4. To calculate the number of moles present you have to divide the amount in grams present by the molar mass of the compound or element.
5. To find the ΔE for one mole, place the ΔE_surroundings over the number of moles present.

Solution

A. ΔE_surr= q

B. The initial temperature is 44.1 °C, the final temperature is 48.4 °C

ΔT= 48.4°C - 44.1 °C

ΔT=4.3 °C

C. C_cal was calculate to be 6.24 kJ/°C and ΔT was calculated to be 4.3 °C. With stoichiometry the °C should cancel out.

q_cal = C_cal • ΔT

q_cal = 6.24 kJ/°C • 4.3 °C

q_cal= 26.83 kJ

-q_cal = ΔE_surroundings

D. The molar mass of oxygen gas is 32.0 g. Given that there is 13.7 g of O₂ we can find the number of moles by dividing the given amount by the molar mass.

\[\dfrac{13.7 g }{32.0 g}\] = 0.428 moles O₂

E. Place ΔE_surroundings (-26.83 kJ) over number of moles present (0.428 moles)

ΔE= \[\dfrac{-26.83 kJ}{0.428 moles}\]

ΔE= -62.69 kJ/mol

Q10.67b

If 0.657 grams of oxalic acid undergoes combustion in a bomb calorimeter, the temperature will increase from 15.7°C to 36.4°C. Find \(\Delta E_{rxn}\) for the combustion of oxalic acid in kJ/mol when the heat capacity of the bomb calorimeter is 4.58 kJ/°C.

Strategy

1. First you need to identity the formula for oxalic acid. Knowing this will help you find the molar mass of the compound, which is the sum of the element's mass that is in the compound.
2. Next, you need to know that \(-q_{cal} = the specific heat \times the difference of the temperature\)
3. When you get the answer to the step above, you divide that answer by the number of moles of oxalic acid.
4. Since the amount in grams of oxalic acid is provided, you need to change it to moles. This is done by: amount of grams given \(\cdot \dfrac{1 mole of oxalic acid}{molar mass of oxalic acid}\cdot\). The number you get will be the number of moles of the solution.

Solution

A. The formula for oxalic acid is C_{2}O_{4}H_{2}. To determine the molar mass you find the atomic mass of Carbon, Oxygen, and Hydrogen and sum them. Since there are two Carbon elements, you multiply the atomic mass of Carbon by two. The same rules follow for Oxygen and Hydrogen.

2(12.01 g of Carbon)+4(15.999 g of Oxygen)+2(1.00784 g of Hydrogen)=90.03 g of C_{2}O_{4}H_{2}

B.-q_{cal}= the specific heat \cdot the difference of the temperature

=4.58 \dfrac{kJ}{°C} \cdot (36.4°C-15.7°C)
=4.58 \dfrac{kJ}{°C} \cdot 20.7°C
=94.81 kJ

Since \(-q_{cal}l=94.81\; kJ\), the 94.81 kJ becomes negative from the -q. Therefore, \(q_{cal}=-94.81\; kJ\).

D. The amount of grams given is 0.657. The number of moles needs to be determined.

moles=0.657 g of C_{2}O_{4}H_{2} \cdot \dfrac{1 mole of C_{2}O_{4}H_{2}}{90.03 g of C_{2}O_{4}H_{2}}
moles= 7.30\cdot 10^{-3} of C_{2}O_{4}H_{2}

C. \Delta E= \dfrac{q_{cal}}{number of moles}
= \dfrac{-94.81 kJ}{7.30\cdot 10^{-3} mol of C_{2}O_{4}H_{2}}
= -1.30\cdot 10^{4} kJ/mol