Rates and equlilbria Flashcards
Rate of reaction definition and equation
the rate of a chem reaction is the change in concentration of a reactants or product per unit if time
reactants used/ time
products produced/time
simple collision theory
for a chem reaction to occur:
molecules must collide
must have greater energy or equal to activation energy
must collide at correct orientation
if one or more of these conditions isn’t meant the no reaction occurs the molecules bounce off one another. An ineffective collision
effect of conc on rate
conc of reactants molecules increases
rate of reaction increases
at higher conc there are more molecules in given volume
more frequent successful collisions
effect of pressure on rate (gases only)
pressure if gas increases and gas molecules pushed closer together
no of gas molecules in given volume increases
more frequent successful collisions occur
same effect as increase conc but that is solids and this is gases
effect of temp on rate
as temp of reaction increases
rate of reaction increases
higher temp the average energy of molecules increases
a greater proportion if molecules have energy greater than or equal to the activation energy
more frequent successful collisions occur
activation energy
is the minimum energy required to start a reaction by breaking of bonds in the reactants
Catalysts and why they are used
increase r of r but not consumed by overall reaction they do this by providing a different pathway for the recation with lower A energy
(if drawing enthalpy profile for this just lower height of curve)
heterogenous- catalyst has diff physical state from the reactants
homogeneous- catalyst and reactants are the same physical state
great economic and environmental benefits-
lower temp
reduce energy demand
less combustion of fossil fuels
and therefore less cost due to lower energy demand and less CO2 emissions
increased sustainability- higher atom economy and less waste
using enzyme also generates specific products no side products
operate close to room temp+ pressur
enzymes commonly in solution with reactants
examples of catalyzed reactions
ammonia( Haber process)- N2 (g) + 3H2 (g) —-> 2NH3 (g) iron is catalyst
polythene- ethene + polythene ziegler-Natta catalyst
catalytic converter( to remove toxic gas)-
2NO (g) + 2CO (g)——–>N2 (g) + 2CO2 (g)
Rh/Pd/Pt (alloy of all 3 metals )
The boltzmann distribution curve
shows the distribution of molecules-
very few have high energy
more have low to mid
x axis- energy
y axis- no of molecules
must start at 0
and can’t be symmetrical
can’t curve up at the end
greatest no of molecules have the mode/average energy, this is peak of curve
total area under curve= total no of molecules
shaded area or area after Ea line is proportion of molecules have > or = Ea
(these can react/ have enough energy)
asymmetrical shape shows more molecules have lower energy
effect of temp- boltzmann curve
higher temp= more molecules have higher energy
average energy increases
curve flattens and shifts right
shaded area/ above Ea area is greater
larger proportion of molecules with > or = Ea
more frequent successful collisions
r of r increases
area under curve= same as same no of molecules
effect of catalyst- boltzmann curve
catalyst increases r of r by lowering Ea
greater proportion of molecules have energy > or = Ea
shaded area/ Area past Ea Is greater
more frequent successful collisions
r of r increases
here the Ea line just move back along curve slightly
procedures to study r of r
reactants used
products formed
measure change in conc of either at regular intervals
diff methods-
measuring Mass of R or P at regular intervals
extracting sample from reaction mixture + analysing by titration at regular intervals (often acid is either R or P
use colourimeter/ spectrophotometer at regular time intervals
measure vol of gas evolved at regular time intervals
measure change in Ph (PH probe) or elec conductivity at reg time intervals
For gases either:
1.vol of gas produced measured at reg time
beaker, reactants, bung + delivery tube and gas syringe
graph increases then levels off
x- time Y-vol of gas
2.measure decreasing mass of R using mass balance at reg time intervals.
beaker, reactants, stopper and balance
graph decreases and levels off
x-time Y-mass
R of R time graphs (conc) + tangents
time graph shows chance in conc of R or P with time
conc- plotted on Y axis from continuous measurements during reaction
can be used to calc R of R at given time
overall rate= gradient of graph over all
total change in Y/ total change in X
initial rate= draw tangent from t=0
change in Y/ change in X
Rate at specific time= draw a tangent at specific time
change In Y/ change in X
if it asks for after Xs just draw target at Xs
when plotting a graph draw smooth curve for line of best fit
Reversible reaction
where both forward and backwards reaction can occur at same time
dynamic equilibrium + how it is reached
exists in a closed system when rate of forward reaction is equal to rate of reverse reaction
doesn’t mean the concentration of P and R are equal but they don’t change
macroscopic properties don’t change e.g. temp/pressure
reaching it:
r of r at beginning of reaction of R is fastest as R have high conc so frequent successful collisions
once sufficient P is made, reverse reaction begins
r of r of reverse starts slow and builds
eventually rate of Forward and backward are =
dynamic equilibrium
can be detected when things like colour or Ph become constant
le chatelier’s principle + equilibrium position
often used to predict the effect of a chage in Temp, pressure or conc of R or P on dynamic equilibrium
Catalyst also effect it
percentage of P to R in a mixture when at dynamic equilibrium shows position of equilibrium
more product = lies to right
more reactants= lies to left
desirable to be as far right as possible for high product yield
effect of conc on dynamic equilibrium
when you increase the conc of one reactants/products
equilibrium shifts to the other side, to minimize the effect of the change in conc
to produce more products/reactants
mention:
where
why
composition
colour change
rember this might be slight cause there doesn’t have to be equal conc of R and P just equal rates of reaction
effects of pressure of dynamic equilibrium
pressure- force exerted by gas particals when they collide with the walls of their containers
only consider gases in system
when pressure increases equilibrium shifts to the side with less moles of gas to minimize the increase in pressure
when pressure decreases
equilibrium shifts to the side with more moles of gas
to minimize the decrease in pressure
if same no of moles then there is no change/effect of equilibrium position
mention?
Where
why
comp
and colour change
effect of tempreatue on dynamic equilibrium
exo- released heat to surrounding
endo- absorb heat
one direction will be endo and the other exo
if the temp is increases
equilibrium will shift to the endo reaction side
to minimize the effect of an increase in temperature by absorbing heat energy
as other reaction is exo (forward or backward reaction)
if temp decreases the opp happens
shifts to exo
to minimize effect of decrease by releasing heat energy
as the other reaction Is endo (foward/backward)
effect of catalyst on dynamic equilibrium
DOESNT AFFECT POSITION OF EQUILIBRIUM
but does increase the rate of dynamic equilibrium
as it increases the rate of both the forward and reverse reactions in equilibrium by the same amount
so unchanging position of equilibrium
equilibrium and industrial processes e.g. Haber process
want to achieve the highest percentage yeild possible of product (far right equilibrium)
also high r of r
sometimes need to compromise between equilibrium and r of r
Fe(s)
N2(g)+3H2(g) <——>2NH3(g) -92kjmol
need to be pure gases to avoid side reactions , lowering percentage yield
N can be obtained from fractional distillation of liquid air
iron catalyst lowers Ea and raises r of r
finely divided for high Sa so more frequent collision and higher r of r
lower temp is best as forward reaction is exo and shift equilibrium to the right
but if too low then r of r is too slow
high pressure is best as shift equilibrium right for higher yield
also increases r of r
but has high running cost plus is dangerous
low pressure - too low too slow
Best comp between rate, yield and safety is 400-500 ⁰c and 200 atmospheres (20000 Kpa)
even with this only 15% of H and N converted rest passes through again
things to remeber when answer why might actual conditions may be diff
High pressure- dangerous + expensive
low pressure+ low temp- too low too slow r of r
high temp- expensive
need a compromise between equilibrium yield + rate of reaction + safety + expense
most talk about r of r and yield
Equilibrium constant, Kc
quatative measure of the proportion (ratio) of product to reactants
Kc indicates where equilibrium position lies in equilibrium
expression = products conc/ reactants conc
R and P are raised to the power of their balancing numbers
units are canceled out to decide the final units, if on the bottom then you need to swap the positives and negative so mol-1 dm3 instead of mol dm-3
Extra=
Kc value depends only on temp
only refers to 1 specific temp
and is unaffected by conc pressure or catalyst at set temp (new equ value will restore og Kc value)
if temp change shifts right- Kc increases
if temp change shifts left-Kc decreases
The significance of the value of Kc/ Kp
the value of Kc indicates the extent of a chemical equilibrium
Kc > 1 equilibrium lies to the right
Kc < 1 equilibrium lies to the left
Kc = 1 equilibrium lies halfway between
if Kc is a large number (e.g. 1000) equilibrium position lies far to the right
and the higher yield of product is achieved
if Kc is small number ( e.g. 1x10-3) equilibrium position lies far to the left
and a lower yield of product is achieved
so the larger Kc value further position of equilibrium lies to right and higher conc of products compared to reactants
The rate equation
Gives the relationship between the rate of reaction and the conc of the reactant (they are raise to the power of their order)
Rate = K[A]^m [B]^n
k= rate constant (larger the value faster reaction, linked to temp)
A= conc of reactant A
B= Conc of reactant B
M= order wrt A
N= order wrt B
what is order of reaction?
Indicates how the conc of the reactant affects r of r
It can have a value of either:
0 = conc of reactant and r of r are independent (no effect)
1= conc of reactant and r of r are directly proportional
2= conc of reactant and r of r have an exponential relationship
Order of reaction graph
Conc = X axis
Rate = Y axis
Order 0 = straight line
Order 1 = diagonal line (directly proportional)
the rate constant can be determined from the gradient of this graph
Order 2 = expontial curve
What happens to rate when the conc of reactants is double at each Order?
Order 0 = no effect (not included in rate equation)
Order 1 = the rate would double as the power doesn’t effect the big no
Order 2 = it would x4 as the 2 would be squared by the power
how to calc overall Order of reaction?
sum of the individual orders
Determining the order of reaction
Pick 2 experiments where only one conc changes
If there is no change in rate = order 0
if there change in conc and rate is portional = order 1
if the conc doubles and the rate quadruples (exponential) = order 2
when answering questions:
state the experiments used
state the change in conc and rate
Then say order wrt
Half-life definition
The half-life of a reactant, t 1/2 is the time taken for the conc of the reactant to reduce by half
How to tell the difference between first and second and 0 order in a conc time graph
0 is a diagonally
The half life of the first order is independent of the conc and is constant
The half life of 5he second order changes dramatically
rate constant for FIRST order=
K= Ln2 / (t1/2)
Rate determining step definition
The rate-determining step is the slowest step of the rate mechanism of a multi step reaction
What does the rate equation tell us about the RDS?
Which species a d how many of them are involved in the RDS of the mechanism
How to write a 2 step mechanism (including RDS)
First look at the rate equation- identify which species are present and in what number
use these in the RDS (this is normally the first step)
aim to produce one of the overall products in the first step
balance charges
Insure that any species not present in the overall equation are cancelled out in the 2 reactions
The number of species in the 2 equations should match with that of the overall equation
The Arrhenius Equation
K= Ae ^-Ea/RT
K=Rate constant
Ea=Activation energy (J mol-1)
R= gas constant
T= temp
A= pre exponential factor (same units as K)
Easier form for Arrehenius
Ln K = -Ea/RT + Ln A
or
Ln K= - (Ea/RT) × (1/T) + Ln A
this follow Y= mx + c
Y= ln K
M = - (Ea/R) (this is the gradient)
X= 1/T
C= Ln A (where line of best crosses X axis)
so A=e^c
Homo vs Hetero equilibrium
Homo= all species have the same state/ phase
Hetero= species have different states /phases
anything solid is removed from Kc only g and Aq
exception= H2O (l) if solvent not aqueous
Calculate Kc from Initial amounts
Step 1:
R- Ratio
I - initial mol
C - change
E - equilibrium mol
× the change by the ratio
Step 2:
calc equilibrium concentration=
equilibrium moles/ equilibrium volume
Step 3: Write the Kc expression and calc units
Kc from experimental techniques example
Analysis by Titration: Example 1
For example, in the following equilibrium, since an acid is involved in the esterfication maction, and an acid catalyst is used, the equilibrium muxture can be analysed using titration with an alkali solution to find the moles of acid present
CH3COOH + C2H5OH –> CH3COOC2H5 + H20
Procedure
- In a conical flask, mix together 0.100 mol CH3COOH and 0.100 mol C2H5OH Add 0.0500 mol of HCl(aq) as an acid catalyst to the flask
The total volume of the mixture in the flask is 20.0 cm³
The amount of water in the aqueous acid catalyst is 0.500 mol - Add 0.0500 mol of HCl(aq) to a second conical flask as a control.
- Stopper both flasks and leave for a week to reach equilibrium. 4. Carry out a titration on the equilibrium mixture using a standard solution of sodium Hydroxide
- Repeat the titration with the control to determine the amount of acid catalyst that has been added.
Results
By analysing the two titrations:
amount of HCl(aq) in control = 0.0500 mol
amount of acid (HCl and CH3COOH) in equilibrium mixture = 0.115 mol amount of CH3COOH in equilibrium mixture 0.115-0.0500=0.065 mol
K. Calculation
CH3COOH C2H5OH CH3COOC2H5 H20
Ratio: 1 1 1 1
Initial: 0100 0.100 0 0.500
Change:-0.035-0.035 +0.035 +0:035
Equ mol: 0.065 0.065 0.035 0.535
Equ conc: 3.25 3.25 1,75 26.75
KC= [CH3COOC2H5] [H2O]
[CHCOOH] [C₂H₂OH]
1.75 x 26.75
3.25 x 3.25
Kc= 4.43 no units
Experimental techniques used to clac Kc
Titration
Ph probe
Colorimeter
Kp basic info
Used for Equ involving gases
in terms of Partial pressure
Kc and Kp are directly related
Kpa Pa or atm
Kp only includes gases - all else ignored
Kp equation
Kp = (p(C)^c × p(D)^d)/ (p(A)^a × p(B)^b)
Partial presseure of products over Partial pressure of reactants
to the power of their balancing number
units= Kpa, Pa, or atm
only effected by temperature
significance is same as KC
Mole fraction equation
No of moles of specific species/ total no of moles in gas mixture
Partial pressure equation
Mole fraction of specific species x total pressure
Calc Kp from initial amounts
1) use RICE to calc equilibrium amounts of all species
2) calc Partial pressure of all species (gas only)
clac mole fractions and × by total equilibrium pressure
3) write Kp expression
Calc Kp from initial amounts
1) use RICE to calc equilibrium amounts of all species
2) calc Partial pressure of all species (gas only)
clac mole fractions and × by total equilibrium pressure
3) write Kp expression
How to explain effect of Temp on Equ pos using Kc/Kp
Exo-
temp increase
equ pos shifts left (as endo and absorbs heat to minimise effect)
New mixture has more reactants and less products
Kc/p will decrease
endo-
temp increase
shift right (forward endo and absorbs heat to minimise effect)
more product and less reactant
Kc/p will increase
Explain effect of conc on equ pos in terms of Kc/p
conc increases
equ pos shifts to side with less moles( in this case products)
(so og Kc/p no longer matches og value)
this minimises increase in conc
conc of products increases
conc of reactants decreases
Kc/p is restores to og value
Explain effect of pressure on equ pos in terms of Kc/p
pressure increased
has a greater effect on the species with highest Partial pressure
in this case product causing numerator to increase
so Kp increases no longer og value
equ pos shifts left (as less moles of reactants in this case)
Partial pressure of products decreases, decreasing numerator
Partial pressure of reactants increases, increasing denominator
Kp is restored
Effect of Catalyst of Kp/c
no effect
increase r of r of both reactions equally
equ pos doesn’t change
so no change in Kc/p
CONSTANT
