Entropy, Free Energy & Reaction Feasibility — Practice · Grade 11 Science
Chapter: Chemical Energy & ThermodynamicsWhat this practice covers
By the end of Entropy, Free Energy & Reaction Feasibility you should be able to:
- I can calculate ΔG with consistent energy units and kelvin temperature.
- I can distinguish thermodynamic favorability from reaction speed.
8 questions, numeric and written answer, each with a worked explanation.
Vocabulary used: entropy, Gibbs free energy, spontaneous.
Problem 1
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
-20 kJ/mol
−50−300(−.100)=−20.
Problem 2
Answer and explanation
Yes. A high activation-energy barrier can limit its rate.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 3
Hints
List the given quantities, select a relationship, and check units before substituting.
Answer and explanation
-3 kJ/mol
12−300(.050)=−3.
Problem 4
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
600 K
30/.050=600.
Problem 5
Hints
Units and absolute temperature must match.
Answer and explanation
−20−300×(−0.040)=−8 kJ/mol.
Problem 6
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
A sufficiently positive entropy change makes TΔS exceed ΔH, producing negative ΔG under appropriate conditions.
Problem 7
Answer and explanation
A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.
The standard reaction free energy ΔG° refers to a defined reference-state account. The actual driving tendency also depends on mixture composition through Q, the reaction quotient, using consistent dimensionless activities. At equilibrium Q=K and the reaction’s ΔG is zero. A negative ΔG° does not mean every possible starting mixture moves forward, because the RT ln Q contribution can change the actual sign. Concentration-only classroom expressions approximate activities and require an explicitly consistent convention.
A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the chemistry relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 8
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.