Chapter 19 Equilibrium Part 2 Flashcards
What is homogeneous equilibrium?
Equilibrium when all the species have the same state/phase
What is heterogeneous equilibrium? How do you write Kc for this and why?
Equilibrium when not all the species have the same state/phase
Ignore solids and liquids in the equation as there concentrations are effectively constant, only use gas/aqueous
How do you determine Kc from original mole values and one equilibrium value?
Look at the moles at the beginning compared to the one that is at equilibrium
The change in moles will be applied to all reactants based on the stoichiometric equation
E.g if 0.4 moles is lost from reactant A which is 2, if reactant B is 1, it will lose 0.2 moles, and if product C is 3 , it will gain 0.6 moles
Then calculate concentration, sub into Kc
What should you do if the volume is not given when trying to calculate KC?
Normally division by volume should cancel out, if 1 to 1 ratio throughout
Or keep as X if unsure, then it should cancel
E.g Esterification
What should you do with catalysts with Kc?
Include them if they are aqueous, or water as a product
This will increase the water value if a product, e.g acid catalyst with Esterification
If no information is given you can assume even with catalyst, starting moles is 0 but needed in the products
What is the mole fraction?
The percentage of moles of a gas mixture that an individual gas contributes
What is partial pressure and the equation?
Partial pressure= mole fraction x total pressure
The contribution of a gas towards total pressure of the mixture
What are the key things to remember with Kp?
Use circular brackets
Only gases
Worked out from mole fraction and partial pressure
You may need to use a table to work out equilibrium moles first
Units- KPa, Atm..
Explain how changing concentrations affect equilibrium using Kc? E.g reactants
If we increase the reactants the reactants, the denominator of Kc is larger, meaning we are dividing by a larger number, so Kc is temporarily too small
However, as KC is constant at a given temperature, the concentration of the products must increase, and the concentration of the reactants must decrease, to restore Kc
The ratio of the reactant to products is now the same, as we have the position of equilibrium to the right
Explain how changing pressure alters the position of equilibrium, using ideas of Kp? Products have more moles than reactants
Increasing the pressure will increase the partial pressure of the products more than the reactants
The numerator is increasing more than the denominator, making Kp temporarily too small
Kp is constant at a given temperature
To restore Kp, the partial pressure of the products must decrease and reactants must increase, pushing the position of equilibrium to the left
How does Kc change with temperature?
If the forward reaction is endothermic, Kc increases when temperatures increases
If the forward reaction is exothermic, Kc decreases when temperature increases
Explain why Kc changes with increasing temperature? The forward reaction is exothermic
Increasing the temperature will increase the rate of the backwards reaction
This increases the concentration of the reactants and decreases the concentration of the products
The numerator is smaller and products increases
This means Kc is smaller and so the position of equilibrium shifts to the left
How do catalysts affect Kc and why?
No effect on Kc
But decreases time needed to reach Kc, as increases the rate of forward and backward reactions by the same factor
How do you practically determine the concentration of organic acid remaining in esterification?
Titrate a sample with the same volume of solely acid catalyst
Then titrate against a sample of the reactant mix
Calculate number of moles of acid in both, and the difference is the number of moles of organic acid in the sample
Find concentration
Do the balancing numbers in Kp matter, as in the multiples in the equation?
Yes, if a different multiple, Kp changes as the units change
You can have Pa to the half, e.g with 1/2 O2
The theoretical number does not, but as the units do, the calculated number does
e.g from Pa to Pa to the 1/2
Square root original Pa value
