Chapter 19 Equilibrium Part 2

Equilibrium when all the species have the same state/phase

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

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

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

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

The percentage of moles of a gas mixture that an individual gas contributes

Partial pressure= mole fraction x total pressure The contribution of a gas towards total pressure of the mixture

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..

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

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

If the forward reaction is endothermic, Kc increases when temperatures increases If the forward reaction is exothermic, Kc decreases when temperature increases

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

No effect on Kc But decreases time needed to reach Kc, as increases the rate of forward and backward reactions by the same factor

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

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

Let pressure= X sub into equation, including with squares, rearrange