Unit 6 Practice: Medium
Thermodynamics · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
Calculate the heat required to warm 100. g of water by 15.0°C.
Use:
c = 4.18 J/g°C
Show answer and explanation
Answer: 6.27 kJ
Work:
q = mcΔT
q = (100.)(4.18)(15.0)
q = 6270 J
Convert:
6.27 kJ
Question 2
A 50.0 g metal sample with c = 0.900 J/g°C cools by 50.0°C. Calculate q.
Show answer and explanation
Answer: −2.25 kJ
Work:
Since it cools:
ΔT = −50.0°C
q = (50.0)(0.900)(−50.0)
q = −2250 J
= −2.25 kJ
Explanation: The negative sign means the metal loses heat.
Question 3
A solution absorbs +2.50 kJ during a reaction. What is q for the reaction?
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Answer: −2.50 kJ
Explanation: If the surroundings gain heat, the reaction loses the same amount.
Therefore:
qreaction = −qsolution
Question 4
A reaction releases 5.00 kJ when 0.0500 mol reacts. Calculate ΔH per mole.
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Answer: −100. kJ/mol
Work:
Because heat is released:
q = −5.00 kJ
Then:
ΔH = −5.00 / 0.0500
= −100. kJ/mol
Question 5
Why does the final temperature of two substances placed in contact usually fall between their initial temperatures?
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Answer: Heat flows from the hotter substance to the colder substance until their temperatures become equal.
Explanation: The hotter substance loses energy while the colder substance gains energy.
They eventually reach thermal equilibrium.
Question 6
Why is the final temperature not always exactly halfway between the two initial temperatures?
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Answer: The substances may have different masses and heat capacities.
Explanation: A substance with greater total heat capacity needs more energy to change its temperature.
Therefore the final temperature is weighted by how much energy each substance can absorb or release.
Question 7
Calculate the heat required to vaporize 2.00 mol of a substance if ΔHvap = 35.0 kJ/mol.
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Answer: 70.0 kJ
Work:
q = nΔHvap
q = (2.00)(35.0)
= 70.0 kJ
Question 8
What happens to temperature during an ideal phase change?
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Answer: It remains constant.
Explanation: During melting or boiling, added energy is used mainly to overcome intermolecular attractions rather than increase particle kinetic energy.
Question 9
What type of energy mainly changes during the flat section of a heating curve?
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Answer: Potential energy
Explanation: During a phase change, particles separate or rearrange.
Their average kinetic energy stays about constant because temperature stays constant.
Question 10
What equation should be used for a sloped section of a heating curve?
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Answer:
q = mcΔT
Explanation: Sloped sections represent temperature changes within one phase.
Question 11
If:
A → B
has:
ΔH = +50 kJ
what is ΔH for:
B → A?
Show answer and explanation
Answer: −50 kJ
Explanation: Reversing a reaction reverses the sign of ΔH.
Question 12
If:
A → B
has:
ΔH = +50 kJ
what is ΔH for:
2A → 2B?
Show answer and explanation
Answer: +100 kJ
Explanation: Doubling the entire reaction doubles the enthalpy change.
Question 13
One reaction step has ΔH = +50 kJ and another has ΔH = −120 kJ. If the steps add to the desired reaction, what is the overall ΔH?
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Answer: −70 kJ
Work:
+50 + (−120) = −70 kJ
Question 14
Why does Hess’s law work?
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Answer: Because enthalpy is a state function.
Explanation: The total enthalpy change depends only on the starting and ending states, not on the route taken between them.
Question 15
Write the formation reaction for 1 mol of H₂O(l) from elements in their standard states.
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Answer:
H₂(g) + 1/2 O₂(g) → H₂O(l)
Explanation: A standard formation reaction must create exactly 1 mol of the compound from its elements in their standard states.
Question 16
Why is the standard enthalpy of formation of diamond not necessarily zero?
Show answer and explanation
Answer: Because diamond is not the standard state of carbon under standard conditions.
Explanation: Graphite is the standard state of carbon.
Only the standard-state form receives:
ΔH°f = 0
Question 17
Calculate ΔH°rxn if:
Products total = −500 kJ
Reactants total = −100 kJ
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Answer: −400 kJ
Work:
ΔH°rxn = products − reactants
= −500 − (−100)
= −400 kJ
Question 18
A reaction requires 600 kJ to break bonds and releases 850 kJ when new bonds form. Estimate ΔH.
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Answer: −250 kJ
Work:
ΔH = broken − formed
= 600 − 850
= −250 kJ
Question 19
Is the reaction in Question 18 exothermic or endothermic?
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Answer: Exothermic
Explanation: Its ΔH is negative.
The energy released from forming bonds exceeds the energy required to break the original bonds.
Question 20
Why are calculations using average bond enthalpies only estimates?
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Answer: Bond energies depend on the exact molecular environment.
Explanation: A C—H bond does not have exactly the same bond energy in every possible molecule.
Tables use average values taken from many substances.