Rates & Equilibrium — Chapter quiz · Physical Science: Chemistry, Grade 11
Distinguish reaction speed from equilibrium composition and calculate direction of adjustment.
Problem 1
- decreasing the concentration of a reactant
- increasing the surface area of a solid reactant
- adding an inhibitor
- lowering the temperature
Answer and explanation
increasing the surface area of a solid reactant
Greater surface area exposes more particles, allowing more effective collisions per second, which speeds the reaction up.
Problem 2
- opposite electric charges
- any amount of energy at all
- enough energy and the correct orientation
- exactly the same mass
Answer and explanation
enough energy and the correct orientation
An effective collision needs at least the activation energy and the proper orientation of the colliding particles.
Problem 3
Answer and explanation
2 mL/s
Average rate = 90 mL ÷ 45 s = 2.0 mL/s.
Problem 4
- so the reaction stops
- in no direction
- toward whichever side is exothermic
- toward the products
- toward the reactants
Answer and explanation
toward the products
By Le Châtelier's principle, the system shifts toward the products to consume the extra reactant.
Problem 5
- all reaction has stopped
- the forward and reverse reactions occur at equal rates
- only the forward reaction still occurs
- reactant and product amounts must be equal
Answer and explanation
the forward and reverse reactions occur at equal rates
Both reactions continue, but at equal rates, so concentrations stay constant — dynamic, not stopped.
Problem 6
Answer and explanation
The kinetic pathway and rate change; overall reaction ΔH and equilibrium K stay the same at fixed conditions. Concentration effects depend on the rate law; catalysts do not change equilibrium yield.
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
The factor is 2²×0.5=2. For total order three, k units are M⁻²s⁻¹. Assume the same law, temperature, medium and relevant pathway remain valid; the change does not establish a new mechanism.
The factor is 2²×0.5=2. For total order three, k units are M⁻²s⁻¹. Assume the same law, temperature, medium and relevant pathway remain valid; the change does not establish a new mechanism.
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
Answer and explanation
Reduced volume raises partial pressures; the adjustment favors the side with fewer gas moles, ammonia, under the model. Equal forward/reverse rates do not imply equal concentrations or that events stop.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 9
Answer and explanation
Q=0.80/0.70≈1.14<4, so net forward conversion is favored. Under the closed 1:1 model total concentration becomes 1.50 M and final values are A=0.30 M, B=1.20 M; the ratio returns, not the original amounts.
Q=0.80/0.70≈1.14<4, so net forward conversion is favored. Under the closed 1:1 model total concentration becomes 1.50 M and final values are A=0.30 M, B=1.20 M; the ratio returns, not the original amounts.
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 10
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
0.08
(.20)²/.50=.08.
Problem 11
Answer and explanation
Temperature can change K; a catalyst alone cannot.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 12
Answer and explanation
Q=0.160/(0.400)²=1.00<2.00, so net forward is favored. From an initial 0.800 M A-only inventory, conservation limits A₂ to at most 0.400 M. The equilibrium value need not reach that maximum, and no rate is given.
Q=0.160/(0.400)²=1.00<2.00, so net forward is favored. From an initial 0.800 M A-only inventory, conservation limits A₂ to at most 0.400 M. The equilibrium value need not reach that maximum, and no rate is given.
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.