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Year 12 chemistry Equilibrium, acids and bases by Mind Map: Year 12 chemistry
Equilibrium, acids and bases
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Year 12 chemistry Equilibrium, acids and bases

Rates of reaction

reaction collision theory

reacting particles must collide

reacting particles must collide with the correct orientation

reacting particles must collide with enough energy to overcome the required activation energy of the reaction

factors affecting rate of reaction

Temperature increases the kinetic energy of reactants so a greater number of particles have the required activation energy needed to react

Concentration increases the number of reacting particles pre volume, so there is a greater number of collisions per second

Surface area: smaller particles have a greater surface area so there is more surface exposed to collide with. = greater number of collisions per second

Catalyst: Lowers the required activation energy of the reaction so a greater number of particles have the required energy to react.

Equilibrium

Le Chateliers Principle

A system in equilibrium will shift to minimise any changes made to it.

Factors affecting equilibrium position

temperature, add heat -equilibrium shifts to the endothermic side to use up heat added

concentration, add reactant - equilibrium shifts to product side to use up added reactant

pressure, increase pressure - equilibrium shifts to side with the least mole of gas to decrease the pressure

Kc = [products]/[reactants]

Reaction enthalpy

Temp changes

E = mc[t2-t1], m=mass of water, c = 4.18kJmol-1g-1, [t2-t1] = change in temp

Reaction enthalpy

If C + O2 --> CO2 H = -341kJmol-1 what mass of C is needed to give off -400kJ of heat?, 1mol of C --> -341kJ ? C --> -400kJ, -400/-341 x 1 = 1.17 mol. 1.17x12 = 14.04g of C is needed

Types of enthalpy changes

exothermic reaction

endothermic reaction

Acids and bases

Bronsted-Lowrey acid

an acid is a H+ donor

a base is a H+ acceptor

Conjugate acid and base pairs

Acid and conjugate bases differ by one H.

bases and conjugate acids differ by only one H.

pH calculations

pH = -log[H3O+]

pOH = -log[OH-]

Kw = [H3O+][OH-], Kw = 1x10-14

Water equilibrium 2H2O --> H3O+ + OH-