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Module 5 summary notes (A Level Chemistry OCR A)
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This document contains summary notes for Module 5 Physical chemistry and transition elements, taken from the A Level Chemistry for OCR A OXFORD textbook.
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MODULE 5PHYSICAL CHEMISTRY AND TRANSMISSION ELEMENTS
CH 18 – RATES OF REACTION
18.1 – ORDERS, RATE EQUATIONS, AND RATE CONSTANTS Rate units =
RATE = QUANTITY REACTED OR PRODUCED
RATE = CHANGE moldm-3 / s
CHANGE IN TIME IN CONC [A] =
CHANGE ‘concentration of
ORDER OF REACTION – The rate of reaction is proportional to the conc of a ∝
rate
reactant raised to a power [A]n
ZERO ORDER – when the conc of a reactant has no effect on the rate, the
reaction is zero order.
A ZERO ORDER REACTION:
~ Any number raised to the power 0 is 1
~ Conc doesn’t influence the rate rate ∝
[A]0
FIRST ORDER – when the rate depends on its conc to the power of 1.
A FIRST ORDER REACTION:
~ conc of A doubles (x 2), reaction rate increases by a factor of 21 = 2
rate ∝
~ conc of A is tripled (x 3), reaction rate increases by a1factor of 31 = 3
[A]
SECOND ORDER – when the rate depends on its conc raised to the power
of 2.
~ conc of A is doubles (x 2), reaction rate increases by a factor of 22 =
4 rate ∝
~ conc of A is tripled (x 3), reaction rate increases by[A]
2
a factor of 32 = 9
Rate = k[A]m[B]n UNITS
Overall order = sum of orders with respect Overall order = 0. Units =
to each reactant moldm-3s-1
Overall order = 1 Units =
INITIAL RATE – the instantaneous rate at s-1
the beginning of an experiment when t=0
18.2 - CONC – TIME GRAPHS
HALF - LIFE – the time taken for half of a reactant to be used up.
~ First order reactions have a constant half-life with the conc halving
every half-life.
~ This pattern is called exponential decay.
CALCULATING RATE CONSTANT FROM THE RATE
Rate = k[A]
So, k = rate / [A]
To find rate: draw tangent and find gradient.
,CALCULATING RATE CONSTANT FROM THE HALF-LIFE
k = ln2 / t1/2
18.3 - RATE – CONCENTRATION GRAPHS AND INITIAL RATES
ZERO ORDER
~ Intercept on the y-axis = rate constant Rate =
k[A]0
~ Reaction rate doesn’t change with increasing conc
So, rate
FIRST ORDER
~ Rate is directly proportional to conc Rate =
k[A]1 line
~ Rate constant = gradient of the straight
So, rate =
SECOND ORDER
~ As this rate-conc graph is a curve, the rate constant cannot be
obtained directly.
~ By plotting a second graph of the rate against conc squared, the
result is a straight line, find=the gradient of this which = k
Rate
k[A]2
THE INITIAL RATES METHOD
Initial rate can be found by measuring the gradient of a tangent drawn at
t=0 on a conc-time graph
18.4 – RATE-DETERMINING STEP
REACTION MECHANISM – series of steps that make up an overall reaction
RATE-DETERMINING STEP – the slowest step in the sequence
18.5 - RATE CONSTANTS AND TEMPERATURE
FACTORS AFFECTING RATE CONSTANT
~ Increasing the temp shifts the Boltzmann distribution to the right,
increasing the proportion of particles that exceed the Ea.
~ As the temperature increases, particles move faster and collide
more frequently.
THE ARRHENIUS EQUATION
K = Ae-Ea/RT
R = gas constant = 8.314Jmol-1K-1
T = temp (K)
A = pre-exponential factor ( frequency factor)
LOGARITHMIC FORM OF THE ARRHENIUS EQUATION
−Ea −Ea 1
ln k = + ln A ln k = + ln A
RT R T
Y = Mx + c
CH - 19 EQUILIBRIUM
19.1 - THE EQUILIBRIUM CONSTANT Kc PART 2
Kc = [products] / [reactants]
,Units: sub units moldm-3 and cancel common units
HOMOGENEOUS EQUILIBRIA – contains equilibrium species that all have
the same state or phase.
HETEROGENEOUS EQUILIBRIA – contains equilibrium species that have
different states or phases.
Kc only includes gases, not solids or liquids
DETERMINING Kc FROM EXPERIMENTAL RESULTS
e.g. A carboxylic acid reacts with an alcohol
1. Mix 0.100mol of the carboxylic acid and 0.100mol of the alcohol
2. Add 0.05mol of HCl (aq) as an acid catalyst to the flask. Total
volume of the mixture is 20cm3
3. The amount of water in the aq acid catalyst is 0.5mol
4. Add 0.05mol of HCl (aq) to a second flask as a control
5. Stopper both flasks and leave for a week to reach equilibrium
6. Carry out a titration on the equilibrium mixture using a standard
solution of sodium hydroxide
7. Repeat titration with the control to determine amount of acid
catalyst that had been added
19.2 - THE EQUILIBRIUM CONSTANT Kp
MOLE FRACTION
~ It’s the same as its proportion by vol to the total col of gases in a
gas mixture.
MOLE FRACTION x(A) = number of moles of A
total n.o of moles in gas mixture
PARTIAL PRESSURE
~ It’s the contribution that the gas makes towards the total pressure
P.
~ The sum of the partial pressures of each gas equals the total
pressure.
P(A) = mole fraction of A x total pressure P
THE EQUILIBRIUM CONSTANT Kp
~ Kp uses partial pressures replacing concentrations in Kc
~ Units: KPa, Pa or atm
~ Kp only includes gases and other species must be ignored
19.3 - CONTROLLING THE POSITION OF EQUILIBRIUM
WHY DOES THE EQUILIBRIUM POSITION SHIFT?
1. If the conc of a species is increased, the equilibrium position shifts
in the direction that reduces the conc.
2. If the pressure is increased, the equilibrium position shifts towards
the side with fewer gaseous moles.
3. If the temp is increased, the equilibrium position shifts in the endo
direction.
, DO EQUILIBRIUM CONSTANTS CHANGE?
K=1 – equilibrium halfway between reactants and products.
K=100 – equilibrium in favour of the products
K=1X10-2 – equilibrium in favour of the reactants
EFFECT OF TEMP ON EQUILIBRIUM CONSTANTS
IF FORWARD REACTION IS EXO:
~ Equilibrium constant decreases with increasing temp
~ Raising the temp decreases equilibrium yield of products
~ Equilibrium shifts to the left
IF FORWARD REACTION IS ENDO:
~ Equilibrium constant increases with increasing temp
~ Raising the temp increases the equilibrium yield of products
~ Equilibrium shifts to the right
The value of k is unaffected by changes in conc or pressure and a catalyst
~ Catalysts effect the rate of a chemical reaction but not the position
of equilibrium
~ Catalysts speed up both the forward and reverse reactions in the
equilibrium by the same factor
CH 20 – ACIDS, BASES AND pH
20.1 - BRONSTED-LOWRY ACIDS AND BASES
~ Acids dissociate and release H+ ions in aqueous solution
~ Alkalis dissociate and release OH- ions in aqueous solution
An alkali is a soluble base.
BRONSTED-LOWRY ACID – a proton donor
BRONSTED-LOWRY BASE – a proton acceptor
CONJUGATE ACID-BASE PAIRS – contains 2 species that can be
interconverted by transfer of a proton.
e.g. dissociation of HCL (aq)
HCL (aq) ⇌ H+ (aq) +Cl- (aq)
~ HCl is a strong acid and the equilibrium position is on the RHS
~ A single arrow can be used to indicate that the forward reaction
effectively goes to completion.
~ In the forwards direction, HCL releases a proton to form its
conjugate base, Cl-
~ In the reverse direction, Cl- accepts a proton to form its conjugate
acid, HCl.
e.g. Combining 2 equations to give acid-base equilibrium Forward direction:
HCl (aq) + OH- (aq) ⇌ H2O (l) + Cl- (aq) HCl is an acid as it
donates H+
Acid 1. Base 2. Acid 2 Base 1 OH- is a base as it
MONOBASIC, DIBASIC AND TRIBASIC ACIDS accepts H+
Refers to the total number of hydrogen ions in the acid that can be direction:
Reverse
replaces per mole H2O is an acid as it
donates H+
Cl- is a base as it
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