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REACTION: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ/mol - Heat added
Equilibrium shifts left (reverse reaction favoured and Nh3 decomposes)
REACTION: 2NO₂(g) ⇌ N₂O₄(g), ΔH = -57 kJ/mol - Volume decreased
Equilibrium shifts right (fewer gas moles on the product side)
REACTION: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) - H₂ removed
Equilibrium shifts right to replace H₂
REACTION: H₂(g) + I₂(g) ⇌ 2HI(g) - Initial concentrations are [H₂] = 0.5 M, [I₂] = 0.5 M, [HI] = 0 M. What happens to Q?
Q < Keq, so the reaction proceeds forward to equilibrium
REACTION: CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) - Effect of adding Na₂CO₃
Equilibrium shifts left due to the common ion effect
REACTION: PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq), Ksp = 7.1×10⁻⁹ - [Pb²⁺] = 1×10⁻⁴ M and [I⁻] = 1×10⁻³ M. Will a precipitate form?
Q = 1×10⁻¹⁰, so no precipitate forms (Q < Ksp)
REACTION: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = -200 kJ/mol - Temperature decreased
Equilibrium shifts right (exothermic reaction favored)
REACTION: CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(g) - Effect of increasing O₂ concentration
Equilibrium shifts right (forward reaction favored)
REACTION: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) - Color observed is pale yellow. What happens if more Fe³⁺ is added?
Equilibrium shifts right, and solution turns redder
REACTION: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) - Effect of adding HCl
Equilibrium shifts left due to increased [Cl⁻] from HCl
REACTION: 2NO(g) + Cl₂(g) ⇌ 2NOCl(g), ΔH = -76 kJ/mol - Pressure increased
Equilibrium shifts right (fewer gas moles on the product side)
REACTION: 2HI(g) ⇌ H₂(g) + I₂(g), Keq = 50 at 500 K - Q is calculated as 70
Reaction shifts left (reverse reaction favored)
REACTION: BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) - Solubility of BaSO₄ in water is 1.0×10⁻⁵ M. What happens if Na₂SO₄ is added?
Solubility decreases due to the common ion effect
REACTION: N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ/mol - Temperature decreased
Equilibrium shifts left (exothermic reverse reaction favored)
WORD PROBLEM: CO₂(g) ⇌ CO(g) + ½O₂(g) - What happens to equilibrium if the pressure is decreased?
Equilibrium shifts right (more gas moles on the product side)
WORD PROBLEM: Pb²⁺ and I⁻ are mixed, and Q is calculated to be 5×10⁻⁸, Ksp for PbI₂ = 7×10⁻⁹ - What happens?
A precipitate forms because Q > Ksp
DELTA H = -100 - Heat added to system
Equilibrium shifts left (favoring the endothermic reverse reaction)
DELTA H = +150 - Heat removed from system
Equilibrium shifts left (favoring the exothermic reverse reaction)
VOLUME INCREASED - Effect on equilibrium
Shifts to the side with more gas moles
VOLUME DECREASED - Effect on equilibrium
Shifts to the side with fewer gas moles
PRESSURE INCREASED - Reaction has unequal gas moles
Shifts to the side with fewer gas moles
PRESSURE DECREASED - Reaction has unequal gas moles
Shifts to the side with more gas moles
CONCENTRATION INCREASE - Reactant added
Equilibrium shifts right (forward reaction favored)
CONCENTRATION DECREASE - Product removed
Equilibrium shifts right (forward reaction favored)
KEQ = 1.01 - Initial Q = 0.50
Reaction proceeds forward to reach equilibrium
KEQ = 0.021 - Initial Q = 0.50
Reaction proceeds in reverse to reach equilibrium
SOLUTION TEMPERATURE INCREASE - Endothermic reaction
Solubility increases as equilibrium shifts right
COMMON ION ADDED - Solubility of ionic compound
Decreases due to the common ion effect
KSP = 1.77×10⁻¹⁰ - Q = 2.5×10⁻⁸
No precipitate forms (Q < Ksp)
KSP = 9.8×10⁻⁹ - Precipitate observed
Q > Ksp (solubility limit exceeded)
OPEN SYSTEM - Effect on dynamic equilibrium
Equilibrium cannot be maintained reaction proceeds continuously
CLOSED SYSTEM - Effect on dynamic equilibrium
Equilibrium is maintained with constant forward and reverse reaction rates
ICE TABLE - Purpose
To calculate equilibrium concentrations and establish Keq or Q values
LE CHATELIER’S PRINCIPLE - Definition
A system at equilibrium will adjust to counteract any imposed change
HABER PROCESS - Optimal conditions
High pressure, moderate temperature, and removal of ammonia favor equilibrium yield.
Note 
Flashcard (69)