Connect equilibrium constraints, representations and evidence
Solve eight unfamiliar equilibrium cases, compare competing models and distinguish calculation agreement from experimental support.
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Infer a central constant without hiding measurement bounds
For hypothetical D(aq) ⇌ 2M(aq), a supplied equilibrium packet at fixed temperature gives [D] = 0.10±0.01 and [M] = 0.30±0.02 mol/L. Ideal activities use c° = 1 mol/L. Enter the central dimensionless K, then explain the endpoint range and why the central value alone does not establish precision.
Quantity / representation
Packet value
D concentration
0.09–0.11 mol/L
M concentration
0.28–0.32 mol/L
Evidence status
Original supplied equilibrium packet; endpoint bounds, not confidence intervals or actual learner measurements
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Cancel a species before taking the square root
At one temperature and compatible standard states, 2X ⇌ Y has K1 = 16 and Y ⇌ Z has K2 = 0.25. X is one conserved unit; Y and Z each contain two. Enter K for X ⇌ ½Z. Explain the transformation of both equations and why the numerical constant is not an additive energy.
Quantity / representation
Packet value
Source 1
2X ⇌ Y; K1 = 16
Source 2
Y ⇌ Z; K2 = 0.25
Target
X ⇌ ½Z
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Allow an exhausted undersaturated endpoint
An original ideal AB(s) ⇌ A+ + B− model has Ksp = 0.0025. Only 0.04 mol solid is added to 2.0 L water with no initial ions. Which final state is supported? Compare the available amount with the saturation requirement and distinguish absent solid from a pure phase still present.
Quantity / representation
Packet value
Supply / volume
0.04 mol AB; 2.0 L
Assigned equilibrium constant
Ksp = 0.0025; c° = 1 mol/L
Model
Fixed temperature and volume; ideal ions; no competing reactions
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Count two common anions per formula unit
A hypothetical M²+(X−)₂ solid dissolves as MX2(s) ⇌ M²+ + 2X−. At equilibrium with solid remaining, an original ideal packet reports molar solubility 0.050 mol/L into a solution initially containing 0.20 mol/L X− and matching nonreacting positive charge. Enter dimensionless Ksp, using c° = 1 mol/L, and explain which concentrations are initial and final.
Quantity / representation
Packet value
Dissolution stoichiometry
One M²+ and two X− per MX2
Initial common ion
[X−] = 0.20 mol/L; no initial M²+
Dissolved amount per volume
0.050 mol/L; excess solid remains
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Interpret total pressure without replacing partial pressure
An ideal gas mixture is initially at chemical equilibrium. Inert gas is added while temperature, volume and reacting-species amounts remain fixed. Total pressure doubles. Which conclusion follows? Explain why knowing only the total pressure change would be insufficient if the volume constraint were missing.
Quantity / representation
Packet value
Held quantities
Temperature and vessel volume
Unchanged immediately
All reacting species amounts
Changed
Inert amount and total pressure
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Use quotient exponents for an unfamiliar compression
For hypothetical 2X(g) ⇌ Y(g), the initial equilibrium has dimensionless K = Q = 9. At fixed temperature, volume is rapidly reduced to one-third before reaction changes the amounts. Enter the immediate Q and predict the subsequent direction. Use the complete quotient rather than a pressure slogan.
Quantity / representation
Packet value
Reaction
2X ⇌ Y; Y contains two X units
Immediate pressure change
Every reacting partial pressure triples
Initial state
Q = K = 9
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Do not infer equilibrium from a stationary open vessel
An open vessel receives reactants continuously and removes product. Two successive concentration measurements agree within their stated resolution. Which conclusion is justified? Explain the difference between chemical equilibrium and an observed steady state with flows.
Quantity / representation
Packet value
Boundary
Open vessel with inlet and outlet
Observation
No resolved concentration change between two readings
Missing evidence
Flow balances, opposing reaction rates and equilibrium quotient comparison
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Keep total pressure fixed during the reaction too
The ideal D(g) ⇌ 2M(g) reference has nD = 1 mol, nM = 2 mol, V = 24 L, RT = 24 L bar/mol, p° = 1 bar and K = 4. Add 10 mol inert gas at fixed 3 bar and the same temperature. A candidate final extent x = 0.50 mol gives nD = 0.50 and nM = 3.00 mol. Enter the final volume in L, then verify total pressure and Q without treating the immediate expanded volume as fixed.
Quantity / representation
Packet value
After addition, before reaction
13 total mol; V = 104 L at 3 bar
Candidate final amounts
D 0.50 mol; M 3.00 mol; inert 10 mol
Held constraint
Total P = 3 bar throughout; temperature fixed
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A reviewer should request fresh work, inspect units, representations and assumptions, and assess explanations separately. Authentic experimental inference, temperature-dependent equilibria, coupled systems and independent practical acceptance remain open. This review samples Unit 7 reasoning and does not certify full-unit mastery.
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