EQUATIONS

Question 1

Equation linking mass, Mr, Moles

Answer 1

mass = Mr x mols

mass (g)

Question 2

Equation linking moles, conc, volume

Answer 2

mols = conc x vol
conc (moldm-3)
vol (dm-3)

Question 3

Equation linking gas volume, moles, 24

Answer 3

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

Question 4

Ideal Gas Equation

Answer 4

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

Question 5

% yield

Answer 5

(actual product amount (mols) / theoretical product amout (mols)) x 100

Question 6

Atom economy

Answer 6

(molecular mass of desired product / sum of all product molecular masses) x 100

Question 7

Bond angles for: linear, trigonal planar, tetrahedral, octahedral

Answer 7

linear = 180
trig. planar = 120
tetrahedral = 109.5
octahedtral = 90

Question 8

Enthalpy Change equation when given mass, temp change, specific heat capacity

Answer 8

q = mc∆t

q = enthalpy change (J)
m = mass (g)
c = specific heat capacity
∆T = temperature change (C)

Question 9

Enthalpy change equation when given bond enthalpy data

Answer 9

Σ(product bond enthalpies) - Σ(reactant bond enthalpies)

Question 10

Enthalpy change equation when given enthalpy of formation data

Answer 10

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

Question 11

Enthalpy change equation when given enthalpy of combustion data

Answer 11

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

Question 12

Rate of Reaction

Answer 12

change in concentration / time

Question 13

Finding Rate Constant from one of those dumb stupiud half life lookin ass graphs

Answer 13

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

Question 14

Equation to Obtain Kc

Answer 14

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

Question 15

mole fraction

Answer 15

number of moles of substance / total number of moles of all substances

Question 16

partial pressure

Answer 16

mole fraction x total pressure (units can be either kPa or Pa or atm doesn’t matter)

Question 17

Kp equation

Answer 17

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

Question 18

Ka equation

Answer 18

Ka = [H+] x [A-] / [HA]

Question 19

What is the rule for strong acids?

Answer 19

Fully dissociates from H+ ions - therefore, [H+] = [HA]

Question 20

What is the rule for weak acids?

Answer 20

Partially dissociates from H+ ions - therefore, [H+] = [A-]

[H+] = √Ka x [HA]

Question 21

Kw equation

Answer 21

Kw = [H+] x [OH-]

Question 22

How do you obtain pKa from Ka?

Answer 22

pKa = -log(Ka)

Question 23

How do you obtain Ka from pKa?

Answer 23

Ka = 10^-pKa

Question 24

How do you obtain pH from [H+]?

Answer 24

pH = -log[H+]

Question 25

How do you obtain [H+] from pH?

Answer 25

[H+] = 10^-pH

Question 26

Entropy equation?

Answer 26

∆S = ΣS(products) - ΣS(reactants)

entropy units = J K-1 mol-1

Question 27

Free Energy equation ?

Answer 27

∆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 !!!

Question 28

How does the ∆G affect feasibility?

Answer 28

• 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.

Question 29

Standard Cell Potential equation

Answer 29

E (cell) = E (positive terminal) - E (negative terminal)