You can start with either the R or P
If you begin with B, you’ll create more A disproportionate to the creation of B because there hasn’t been established enough A to start synthesizing into B, thus making the creation of A much faster than the creation of B (and vice versa if you began with A)
The pressure-equilibrium constant
atmospheres
Kp=Kc((RT)^Δn)
Δn=ΣP(being the Σ of the coefficients)-ΣR(being the Σ of the coefficients)
The R stands for the universal gas constant: 0.08206 Latm/molK
The T stands for absolute temperature (k)
Make an ICE (initial, change, equilibrium) chart are use reaction stoichiometry to find the other []s at equilibrium using algebra
Since this is stoichiometrically based, you must multiply the concentration by 2
This is based off the idea that depending on your
Predict the direction of a reaction (refer back to the magnitude of K)
Calculate the amounts of R and P once equilibrium has been reached if we know the initial amounts of at least one substance and have the volume (use total volume when solving at equilibrium if the problem given accounts for additive volume changes)
The system isn’t at equilibrium, which goes for Q<K too
When Q=K, your system is at equilibrium
You need more reactants
You need more products
In this diagram, we added excess H2, disturbing the ratio of concentrations, forcing the reaction to shift RIGHT to a new equilibrium (making more P than R)
If NH3 is removed, the equation will still shift RIGHT to accommodate for the lack of NH3, as then we would still have a disproportionate amount of R over P
N2 goes down to accommodate for a new ratio of more NH3 being produced (almost like N2+4H2⇌3NH3)
H2 is added at the spike, and equilibrium has begun to establish as the elements levels plateau after the spike
In exothermic reactions (∆H=-x):
As temperature rises:
As temperature decreases:
Your equation shifts left and K decreases (think of heat as a product with exothermic reactions and when you have more P, you create more R, thus making a larger denominator and thus lowering K)
Your equation shifts right and K increases
In endothermic reactions (∆H=+x):
As temperature rises:
As temperature decreases:
Your equation shifts right and K increases
Your equation shifts left and K decreases
The forward and reverse reaction rates increase PROPORTIONALLY, thus there is NO SHIFT
Catalysts really only change the reaction rate, but the final composition and K are unchanged
Acids increase [H+] (protons) and bases increase [OH-] WHEN AQUEOUS IN WATER
They’re limited to aqueous solutions
Acids can transfer H+ to other substances (proton DONORS) and bases can accept H+ from other substances (proton acceptors)
H+ and H3O+ (H3O+ just being when H2O accepts an H+ ion)
It’s NOT limited to aqueous solutions
A substance that can act as either a base or acid depending on what its being reacted with and the RXN conditions
H2O, which if added to an acid can act as a weak base, and vice versa with its addition to a base where it acts as a weak acid
They differ by the base losing a H+ (if its the first dissociation/ionization), and the acid has an extra H+
HNO2 (acid) turning into NO2- (conjugate base)
H2O (Amphoteric base) turning into H3O+ (conjugate acid)
The auto-ionization of water
H2O(l)+H2O(l)⇌H3O+(aq)+OH-(aq)
Kw=[((H3O)^+)][((OH)^-)]=[((H)^+)][((OH)^-)]=(1.0*10)^(-14)
25°C
both H3O+ and H+ are interchangeable, so you can use either side
You have an acid when [H+]>[OH-]
You have a base when [H+]<[OH-]
You have a neutral system when [H+]=[OH-]
pH decreases
The pH scale goes from 0-14, 0 being completely acidic and 14 being completely basic, 7 being neutral with equal concentrations of [H+] and [OH-]
Find your [H+] with the Kw equation
Plug the new [H+] into the pH log equation
They exist entirely as dissociated ions
Due to the fact they exist as dissociate ions, they can’t be at equilibrium
Since they can’t reach equilibrium, they’re always drawn with a one-sided arrow
Oxides contain O^(2-)
Hydrides contain H^-
Nitrides contain N^(3-)
Most acids are weak acids
This essentially means they only PARTIALLY IONIZE, making equilibrium, and are difficult to separate, meaning their IMFs are strong
your acid [] decreases (and vice versa)
The more your acid ionizes, the more P you produce, thus taking away from your original acids []
They have more than one ionizable H+
Sulfurous acid, H2SO3
K(sub b(for base))=([conj. acid]*[OH^(-)])/[weak base]
To find the Kb for WEAK bases
You can replace the “H” in pH with pKa or pKb and solve as you would for pH or pOH
pKa=-logKa or pKb=-logKb and pKa+pKb=14 (at 25ºC)
Hydrolysis occurs when dissociated salt ions split the H2O in solution into H+ and OH-, or more specifically when the ACETATE ion (derived from SODIUM ACETATE) reacts with H2O to produce ACETIC ACID and the HYDROXIDE ION
OAc^(-)+H2O⇌HOAc+OH^(-)
The anion of a weak acid (or its conjugate base) reacts with H2O as a base to form OH- and go back to its acidic state, thus making a basic solution (or the anion, X- acts as the proton ACCEPTOR thus acting as a base)
Since the acid is weak, it has strong enough IMFs to attract the H+ from the H2O back and become an acid again