Unit 6 Practice: Hard
Thermodynamics · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
A hot metal sample is placed into cooler water in an insulated calorimeter. Write the energy-conservation equation.
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Answer:
qmetal + qwater = 0
or:
qmetal = −qwater
Explanation: If no significant energy escapes to the surroundings, whatever energy the metal loses is gained by the water.
Question 2
Explain the energy transfer from a hot metal to colder water at the particle level.
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Answer: Faster-moving particles in the metal transfer kinetic energy through collisions to slower-moving water particles.
Explanation: The metal begins with a higher temperature and therefore greater average particle kinetic energy.
Energy transfers until both materials reach the same final temperature.
Question 3
A 75.0 g metal at 95.0°C is placed into 100.0 g water at 20.0°C. The final temperature is 25.0°C. Find the metal’s specific heat.
Use:
cwater = 4.18 J/g°C
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Answer: About 0.398 J/g°C
Work:
Water gains heat:
qwater = mcΔT
qwater = (100.0)(4.18)(25.0−20.0)
= 2090 J
Metal loses same amount:
qmetal = −2090 J
For metal:
q = mcΔT
−2090 = (75.0)(c)(25.0−95.0)
−2090 = (75.0)(c)(−70.0)
c = 2090 / 5250
≈ 0.398 J/g°C
Question 4
A reaction warms 150. g of solution from 22.0°C to 29.0°C. Assume c = 4.18 J/g°C. Estimate qreaction.
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Answer: −4.39 kJ
Work:
Solution:
qsolution = mcΔT
= (150.)(4.18)(7.0)
= 4389 J
= +4.39 kJ
Therefore:
qreaction = −4.39 kJ
Explanation: The solution warms, so the solution gains heat.
That heat must have been released by the reaction.
Question 5
In Question 4, suppose 0.0250 mol of reaction occurred. Calculate ΔHrxn per mole.
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Answer: −176 kJ/mol
Work:
ΔH = −4.39 kJ / 0.0250 mol
= −175.6 kJ/mol
≈ −176 kJ/mol
Question 6
A bomb calorimeter has a total heat capacity of 8.50 kJ/°C. A reaction raises the temperature by 2.40°C. Calculate qreaction.
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Answer: −20.4 kJ
Work:
qcal = CcalΔT
= (8.50)(2.40)
= +20.4 kJ
Therefore:
qreaction = −20.4 kJ
Explanation: The calorimeter gains energy, so the reaction loses it.
Question 7
Why does a bomb calorimeter measure ΔE more directly than ΔH?
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Answer: Because a bomb calorimeter operates at constant volume.
Explanation: At constant volume, expansion work is limited.
Therefore:
qv = ΔE
In contrast, coffee-cup calorimetry is usually performed at constant pressure, where:
qp = ΔH
Question 8
A substance is heated from solid below its melting point to gas above its boiling point. What types of calculations may be required?
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Answer: Several q = mcΔT calculations plus phase-change calculations using q = nΔHphase.
Explanation: A full heating curve can include:
heating solid
melting
heating liquid
vaporizing
heating gas
Each segment must be calculated separately and then added.
Question 9
Why is ΔHvap usually larger than ΔHfus?
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Answer: Vaporization requires much greater separation of particles than melting.
Explanation: During melting, particles remain fairly close together in the liquid.
During vaporization, they separate much more completely, so more intermolecular attraction must be overcome.
Question 10
A heating curve has a long boiling plateau but a short melting plateau. What does that suggest about ΔHvap compared with ΔHfus, assuming the same heating rate and amount of substance?
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Answer: ΔHvap is larger.
Explanation: A longer plateau means more energy is required while temperature remains constant.
That indicates a greater enthalpy change for the phase transition.
Question 11
Use Hess’s law.
Given:
C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ
CO(g) + 1/2 O₂(g) → CO₂(g) ΔH = −283 kJ
Find ΔH for:
C(s) + 1/2 O₂(g) → CO(g)
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Answer: −111 kJ
Work:
Reverse the second reaction:
CO₂ → CO + 1/2 O₂
ΔH = +283 kJ
Add to first reaction:
C + O₂ → CO₂ CO₂ → CO + 1/2 O₂
Cancel CO₂:
C + 1/2 O₂ → CO
Add enthalpies:
−394 + 283 = −111 kJ
Question 12
In a Hess’s law problem, what happens to ΔH if you multiply an equation by 3?
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Answer: Multiply ΔH by 3.
Explanation: Enthalpy is an extensive quantity.
Tripling the amount of reaction triples the energy change.
Question 13
Calculate ΔH°rxn for:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Given:
ΔH°f CH₄ = −74.8 kJ/mol
ΔH°f CO₂ = −393.5 kJ/mol
ΔH°f H₂O(l) = −285.8 kJ/mol
ΔH°f O₂ = 0
Show answer and explanation
Answer: About −890 kJ/mol
Work:
Products:
−393.5 + 2(−285.8)
= −965.1 kJ
Reactants:
−74.8 + 0
= −74.8 kJ
Then:
ΔH°rxn = −965.1 − (−74.8)
= −890.3 kJ/mol
Question 14
Why must physical states be included when using standard enthalpies of formation?
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Answer: Different physical states can have different enthalpies.
Explanation: For example:
H₂O(l) and H₂O(g) do not have the same ΔH°f.
The difference reflects the energy required for vaporization.
Question 15
A student calculates reaction enthalpy using bond enthalpies and gets a value slightly different from the tabulated ΔH°rxn. Why is that expected?
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Answer: Bond enthalpies are average gas-phase bond energies, so they are approximate.
Explanation: Standard formation enthalpies are specific to particular substances and states.
Average bond energies do not capture every molecular environment exactly.
Question 16
Estimate ΔH for:
H₂ + Cl₂ → 2HCl
Given average bond enthalpies:
H—H = 436 kJ/mol Cl—Cl = 243 kJ/mol H—Cl = 431 kJ/mol
Show answer and explanation
Answer: −183 kJ
Work:
Bonds broken:
436 + 243 = 679 kJ
Bonds formed:
2(431) = 862 kJ
Then:
ΔH ≈ 679 − 862
= −183 kJ
Question 17
Why does forming stronger bonds tend to make a reaction more exothermic?
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Answer: Stronger bond formation releases more energy.
Explanation: If the bonds formed in the products are much stronger than the bonds broken in the reactants, more energy is released than absorbed.
This makes ΔH more negative.
Question 18
A student says, “A reaction with a very negative ΔH must be fast.” Explain why this is wrong.
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Answer: Reaction rate depends on activation energy, not directly on ΔH.
Explanation: ΔH compares the energies of reactants and products.
Activation energy describes the barrier between them.
A very exothermic reaction can still be slow if its activation energy is large.
Question 19
A student says, “Breaking bonds releases energy because atoms become free.” Explain the error.
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Answer: Breaking bonds always requires energy input.
Explanation: Bonded atoms attract one another.
Energy must be supplied to separate them.
Energy is released when new bonds form.
Question 20
Explain how energy conservation connects calorimetry, Hess’s law, and bond enthalpy calculations.
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Answer: All three rely on tracking energy changes without creating or destroying energy.
Explanation: In calorimetry:
energy lost = energy gained
In Hess’s law:
individual reaction energy changes add to the total energy change.
In bond enthalpy calculations:
energy required to break bonds is compared with energy released when new bonds form.
All of these are applications of conservation of energy.