Equilibrium Constants & Reaction Quotients — Practice · Grade 11 Science
Chapter: Rates & EquilibriumWhat this practice covers
By the end of Equilibrium Constants & Reaction Quotients you should be able to:
- I can form K and Q expressions with phases and coefficients considered.
- I can predict net adjustment and solve a simple equilibrium composition.
8 questions, numeric and written answer, each with a worked explanation.
Vocabulary used: reaction quotient Q, ICE table.
Problem 1
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
3
K=.60/.20.
Problem 2
Answer and explanation
Its activity is approximately constant while that pure phase is present.
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
2
.80/.40=2.
Problem 4
Answer and explanation
Net reverse change lowers the product-to-reactant expression toward K.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 5
Hints
Compare Q and K for the same expression and temperature.
Answer and explanation
Reverse toward A, reducing Q toward K.
Problem 6
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
Their activities are treated as unity when the pure phase is present; changing its amount alone does not change that activity.
Problem 7
Answer and explanation
Each A₂ requires two A units, so producing more than 0.500 M A₂ would require a negative remaining A amount.
For dimerization, producing x concentration units of A₂ consumes 2x of A. The ICE table therefore differs from a one-to-one isomerization. Substitution into the selected K equation can yield more than one algebraic root; reject any root that makes a species concentration negative or exceeds available material. The displayed K=2 case gives x≈0.3048 M and [A]≈0.3904 M. Substitute the accepted values back into both the equilibrium ratio and atom-equivalent conservation to check the result.
Each A₂ requires two A units, so producing more than 0.500 M A₂ would require a negative remaining A amount.
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
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.