Chemical Energy & Thermodynamics — Chapter quiz · Physical Science: Chemistry, Grade 11
Use calorimetry, Hess’s law, entropy, and free energy with explicit systems and signs.
Problem 1
- negative
- always zero
- always exactly 1
- positive
Answer and explanation
negative
An exothermic reaction releases energy to its surroundings, so the system loses energy and ΔH is negative.
Problem 2
Answer and explanation
10450 J
ΔT = 45.0 − 20.0 = 25.0°C, so q = mcΔT = 100 × 4.18 × 25 = 10,450 J.
Problem 3
Answer and explanation
ΔH=500−650=−150kJ for the represented amount; exothermic. Average bond values are approximate. Breaking a bond costs energy; the overall reaction can release more when product bonds form.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 4
Answer and explanation
ΔH≈720−880=−160 kJ for the selected equation amount. Doubling all coefficients gives approximately −320 kJ. Average bond energies are an approximate full-reaction estimate, not proof about spontaneity or rate.
ΔH≈720−880=−160 kJ for the selected equation amount. Doubling all coefficients gives approximately −320 kJ. Average bond energies are an approximate full-reaction estimate, not proof about spontaneity or rate.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 5
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
-12 kJ
−20+8=−12.
Problem 6
Answer and explanation
Actual bond environments and phases differ from the tabulated averages.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 7
Answer and explanation
qwater=50.0×4.18×8.00=1672 J. qreaction≈−1.672 kJ. Dividing by 0.0200 mol gives approximately −83.6 kJ/mol on the specified reacting-amount basis, subject to the neglected terms.
qwater=50.0×4.18×8.00=1672 J. qreaction≈−1.672 kJ. Dividing by 0.0200 mol gives approximately −83.6 kJ/mol on the specified reacting-amount basis, subject to the neglected terms.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 8
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
2 kJ/mol
10−8=2.
Problem 9
Answer and explanation
No. It changes reaction rate pathways, not the overall state-function difference.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 10
Answer and explanation
At 250 K, ΔG=30.0−250(0.150)=−7.50 kJ/mol. The modeled crossover is 30.0/0.150=200 K. Actual ΔH/ΔS can vary with temperature and phases, and the calculation does not establish a rate.
At 250 K, ΔG=30.0−250(0.150)=−7.50 kJ/mol. The modeled crossover is 30.0/0.150=200 K. Actual ΔH/ΔS can vary with temperature and phases, and the calculation does not establish a rate.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.