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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.

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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?

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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?

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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?

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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.