chemical reactivity Flashcards
what is a conjugate pair
base formed when an acid donates a proton (H+) and an acid formed when a base accepts a proton (H+)
strong acids
HCl, H2SO4, HNO3
weak acids
CH3COOH, NH4+, HF
Strong bases
carbonates, hydroxides, metal oxides, hydrogen carbonates, NaOH, KOH
weak bases
NH3
amphiprotic
can accept or donate H+ (act as acid or base depending on the other substance)
strong acid qualities
completely dissociates to form high conc. of H3O+
low ph as pH depends on conc. of H+ as H3O+
only forward reaction
high conduction of electricity due to lots of H3O+ ions
no strong acid left
weak acid qualities
partially dissociates in water so only some H+ ions break away from molecule
lower pH than strong acid
conducts electricity but weakly as there are still H3O+ ions being produced
reversible reaction - dynamic equilibrium
most weak acid left
strong base qualities
fully dissociate in water to produce lots of OH- ions and greater proportion of the reactant readily accepts proton
conducts electricity well due to large OH- being produced
one way forward reaction
High ph due to lots of OH- ions present therefore only a few H3O+ ions due to water self ionizing
no strong base left
weak base
partially dissociate and smaller proportion of reactants convert to products by only some accepting protons
produces less OH- ions but still mainly OH- therefore pH is high but lower than strong bases
dynamic equilibrium
conducts electricity still but weaker due to less OH- charged particles freely moving
most weak base left
acid + base —->
salt + water
ionic salt reaction
split salt to the two different ions
then weak acid or base forms and react with water = dynamic equilibrium and the left over ion from salt becomes spectator
state the pH and why and whether it can conduct electricity
can conduct because although few H3O+ ions or OH- ions are being produced there is still a high conc of ions from the initial ionic salt split
collision theory
particles must collide - and with correct orientation and with sufficient activation energy
way of writing collision theory
state the change
how this affects collision theory
increase ROR or decrease
for temp - two things = kinetic energy and less or greater proportion
catalyst added = alternative pathway with less activation energy so greater proportion overcome activation energy barrier for successful collisions per second
conc = per unit volume
surface area = more or less particles exposed
draw exo and endo graphs
Kc =
[products]/[reactants]. only gas
if Kc greater than 1 = reaction favours products
Kc less than 1 = reaction favours reactants
Kc = 1 then neither reactants or products favoured as both are in equal ratio
Kc is equilibrium constant
Q
any concentration of reactant and products at any time to compare with Kc and see if system is at equilibrium
le chatellier
system will be at equilibrium unless a change is made, then the system will do the opposite of the change made to minimise this change and reestablish equilibrium
write dynamic equilibrium
state change, how will equilibrium respond, what direction favoured and why (remember to explain this)
colour
observation straight after change not compared to beginning
increase pressure
only gases
more particles hitting sides of container to minimise this less moles direction favoured so pressure decreases as less particles hit container so product conc increases or decreases depending on this
decrease pressure
less particles hitting container so more moles direction favoured so more hit container and increase pressure
same moles
pressure doesn’t affect as there are equal mole ratio
writing pressure
state change - how is this minimised - which direction - why - what does this produce more of
endo
absorb heat
surrounding temp decreases
favour exo to increase heat
exo
releases heat and surrounding temp increases so favour endo to decrease heat
rH
forward reaction
change in temp = change in Kc value
temperature affects K because changes in temperature shift the reaction toward a certain side. Think of temp. as a reactant or product in the reaction. If endothermic, it would look something like this:
Reactants + Temperature -> Products
If you increase the temperature, the reaction will shift toward the products in equilibrium, and hence Kc would be have to be changed.
because a certain direction is exo or endo to produce products and is in closed system so temp cant escape
catalyst
increases the rate of forward and reverse reaction at the same time so equilibrium does not change but catalyst makes a system reach equilibrium faster
formula for OH- conc
[OH-] = 10^-(14-pH)
