EQUATIONS Flashcards
Equation linking mass, Mr, Moles
mass = Mr x mols
mass (g)
Equation linking moles, conc, volume
mols = conc x vol
conc (moldm-3)
vol (dm-3)
Equation linking gas volume, moles, 24
vol = mols x 24
vol (dm-3)
This is based off the concept that 1 mole of a gas takes up 24dm-3 of space
Ideal Gas Equation
pV = nRT
p = pressure (Pa) V = volume (m3) - ref. cm3 -> dm3 = divide by 1000 dm3 -> m3 = divide by 1000 n = moles R = 8.314 T = temp (Kelvin) - ref. 0C = 273K
% yield
(actual product amount (mols) / theoretical product amout (mols)) x 100
Atom economy
(molecular mass of desired product / sum of all product molecular masses) x 100
Bond angles for: linear, trigonal planar, tetrahedral, octahedral
linear = 180
trig. planar = 120
tetrahedral = 109.5
octahedtral = 90
Enthalpy Change equation when given mass, temp change, specific heat capacity
q = mc∆t
q = enthalpy change (J) m = mass (g) c = specific heat capacity ∆T = temperature change (C)
Enthalpy change equation when given bond enthalpy data
Σ(product bond enthalpies) - Σ(reactant bond enthalpies)
Enthalpy change equation when given enthalpy of formation data
Σ(enthalpy of formation of products) - Σ(enthalpy of formation of reactants)
alternatively, draw Hess cycle that has the arrows going from the products to the reactants / products
Enthalpy change equation when given enthalpy of combustion data
Σ(enthalpy of combustion of reactants) - Σ(enthalpy of combustion of products)
alternatively, draw a Hess cycle that has the arrows going down from the reactants/products to the combustion products on the bottom
Rate of Reaction
change in concentration / time
Finding Rate Constant from one of those dumb stupiud half life lookin ass graphs
k = ln(2) / half life
ln(2) is not a value you obtain; its literally just on the calculator oh my god please dont make this mistake
Equation to Obtain Kc
Kc = [C]^c x [D]^d / [A]^a x [B]^b
for reaction aA + bB -> cC + dD
(the arrow is an equilibrium arrow btw)
keep in mind that if the equation is heterogeneous then include ONLY GASES in Kc equation - always look at the mf state symbols
mole fraction
number of moles of substance / total number of moles of all substances
partial pressure
mole fraction x total pressure (units can be either kPa or Pa or atm doesn’t matter)
Kp equation
Kp = (PC)^c x (PD)^d / (PA)^a x (PB)^b
for reaction aA + bB -> cC + dD
the arrow is an equilibrium arrow btw
P = partial pressure
Ka equation
Ka = [H+] x [A-] / [HA]
What is the rule for strong acids?
Fully dissociates from H+ ions - therefore, [H+] = [HA]
What is the rule for weak acids?
Partially dissociates from H+ ions - therefore, [H+] = [A-]
[H+] = √Ka x [HA]
Kw equation
Kw = [H+] x [OH-]
How do you obtain pKa from Ka?
pKa = -log(Ka)
How do you obtain Ka from pKa?
Ka = 10^-pKa
How do you obtain pH from [H+]?
pH = -log[H+]
How do you obtain [H+] from pH?
[H+] = 10^-pH
Entropy equation?
∆S = ΣS(products) - ΣS(reactants)
entropy units = J K-1 mol-1
Free Energy equation ?
∆G = ∆H - T∆S
- T (temp) in Kelvin units for ∆S and ∆H have to be the SAME - ∆H usually kJ mol-1 - ∆S usually J K-1 mol-1 usually turn ∆S to kJ - Divide the ∆S value by 1000 !!!
How does the ∆G affect feasibility?
- If ∆G is always negative, in the case of a negative ∆H and a positive ∆S, then reaction is always feasible
- If ∆G is always positive, in the case of a positive ∆H and a negative ∆S, then reaction is never feasible.
However, if at low temperatures both ∆H and ∆S are negative, then reaction is feasible at low temperatures
Similarly, if at high temperatures both ∆H and ∆S are positve, then reaction is feasible at high temperatures.
Standard Cell Potential equation
E (cell) = E (positive terminal) - E (negative terminal)
