Unit 7 - Equilibrium Flashcards
Define equilibrium
The equilibrium state is dynamic, meaning that both the forward and reverse reactions continue to occur at equal rates, resulting in no net observable change in the system
Reactants and products are both simultaneously present
The concentrations and the partial pressures of the products and reactants stop changing (remain constant), but this does NOT mean they are equal
Equilibrium for evaporation and dissolution
Initially, the rate of evaporation is much larger than the rate of condensation. Eventually, dynamic equilibrium is achieved at the surface of the water when rate of evaporation = rate of condensation
Initially, the rate of dissolution of a solid in a solvent is much larger than the rate of precipitation. Eventually, dynamic equilibrium is achieved in a saturated solution, when rate of dissolution = rate of precipitation
Equilibrium graph
Equilibrium is reached when the concentration curves flatten out, that is, when the concentrations of the products and reactants stop changing
The coefficients in the balanced equation tell you how much the concentration of each species will change from their initial concentration to their equilibrium concentration. This is witnessed by proportional differences in the height of the curve
EX: 3 H2 + N2 → 2 NH3
Concentration of H2 will decrease three times as much as concentration of N2
What happens if the denominator of the reaction quotient (Q) is 0?
Q will approach infinity
Q > K
the reaction will shift left or in the reverse direction to produce more reactant
How do you represent a multistep reaction with different K values for each individual step?
If you reverse a reaction, take the reciprocal of the original equilibrium constant to get the new constant (Kforward x Kreverse = 1)
If you multiply a reaction by a coefficient, take the original equilibrium constant to that power to get the new constant
EX: N2 + 3 H2 → 2 NH3 turns into 1/2 N2 + 3/2 H2 → NH3
Since the equation was multiplied by 1/2, Keq will be raised to the power of 1/2
If you add two reactions together, multiply the equilibrium constants of those reactions to get the new constant
Kc vs Kp
Kc = molar concentrations of aqueous ions Kp = partial pressures of gases
*don’t include solids or pure liquids
Ksp (definition, reaction, applications)
Ksp = solubility product for a given solute
Reaction - solid ionic compound dissolves into ions (saturated solution)
EX: AgI (s) ⇌ Ag+ (aq) + I- (aq) Ksp = [Ag+] [I-]
Ksp has no denominator because the reactant is solid
The greater the value of Ksp, the more soluble the salt (greater concentrations of ions) – ONLY applies to comparison of salts that all have a 1:1 ratio of ions (EX: Works for AgCl vs AgBr but not for Mn(OH)2 vs MnO)
The lower the value of Ksp, the less soluble the salt, so it will precipitate first and have a smaller percent of dissociation/ionization
Ka and Kb
Ka = acid dissociation constant for weak acids
Kb = base dissociation constant for weak bases
The smaller the value of Ka or Kb, the weaker the acid or base (respectively) because that means there is greater concentration or reactant which shows low dissociation
Keq
General equilibrium constant
A larger value of Keq means a higher concentration or pressure of products at equilibrium (reaction will be proceed almost entirely to completion)
*can’t be zero or negative
Reaction quotient Q
Writes the equilibrium constant with the initial concentrations of reactants and products
Compare with Keq to determine which direction the reaction proceeds in order to reach equilibrium
If Q = K, equilibrium, no shift in either direction
If Q > K, shift to the left (reactants side, reverse)
If Q < K, shift to the right (products side, forward)
How do you solve equilibrium problems?
- if the value of K is very large, you can assume the reaction goes to completion and just solve the problem with molar coefficients and limiting reactants
1. Write the balanced equation for the reaction, including all states
2. Write the equilibrium expression (be careful to note whether it’s Kp or Kc).
3. Then use the given initial concentrations to calculate Q. Determine the direction of the shift to equilibrium
4. Make an ICE table. Define the change needed to reach equilibrium and apply it to the initial concentrations to find the equilibrium concentrations (change is dependent on molar coefficients)
5. Substitute the equilibrium concentrations into the equilibrium expression with K (which is given) and solve for x.
If both the numerator and denominator are squared, take the square root of both sides
- Write all the numerical values for the equilibrium concentrations underneath the ICE table (substitute in x)
What is an external stress on equilibrium? How does it affect equilibrium position?
A stress is a change in pressure, temperature, or concentration
When the equilibrium position shifts due to changing concentrations or pressure/volume, it is temporary, meaning the reaction will eventually reestablish the same ratio of products to reactants it had at equilibrium. Keq stays constant.
However, temperature changes will cause a shift AND change Keq for the reaction
The student conducted two experiments, one with a reactant mass of 50 g and another with 100 g. The final pressure of the product stayed the same in both cases. Explain why.
If you double the reactant mass or concentration, you would expect the product concentration and partial pressure to also double. If this doesn’t happen, it means the system has reached equilibrium (the reactant has not fully reacted).
How do you solve a problem with Ksp?
- Write the balanced equation for the reaction, including all states
- Write the expression for Ksp. (It will be the product of the concentrations of the aqueous ions)
- Make an ICE table. Define the change needed to reach equilibrium and apply it to the initial concentrations to find the equilibrium concentrations (change is dependent on molar coefficients)
- Substitute the equilibrium concentrations into the equilibrium expression with Ksp value given and solve for x, or the solubility of the compound.
You can also solve for an unknown Ksp value by substituting x, or solubility, into the equilibrium constant expression
Common ion effect
The solubility of an ionic solid (salt) is reduced when it is dissolved into a solution that already contains one of the ions present in the ionic solid (salt).
EX: Adding NaF to a solution that already contains SrF2 (dissociates readily into Sr2+ and F- ions) will cause it to have a lower percent dissociation.
This is because of Le Chatelier’s principle. By adding greater concentration of product (ions), equilibrium position will shift to the left in the direction of the reverse reaction, toward the reactants side. This will decrease the solubility of the salt
