# chemistry rates equilibria, rate and pH

## rates of reaction

Equation

Rate= change in concentartion ofreactant or product/time taken for cahnge to take palce

units= MolDM-3/s= MolDM^-3s^-1

information

use square brackets to represent the concentration of the chemicals

e.g

[CO2] means concentration of CO2 in Mols DM^3

rates and graphs

as the reaction proceeds it will palto this is when the graph has reached its maximum rate as the reactants are used up the rate will begin to deacrese evtualy coming to an end point

1 of 29

## orders

Orders are the effect that different chemicals have on the rate of reaction

• Three orders
• Zero order
• First order
• Second order

Zero order effects

• They have no overall effect on the rate of reaction due to their addition as a reactant
• However they can be catalysis’s so though as a reactant they don’t effect the rate they do effect it as a catalysts
• They however do not affect the rate equation

First order

• A first order is written like so [A] ^1
• As you increase [A] ^1 you increase the rate by the same factor
• E.g. you increase the amount of [A] to 2 [A] as such the rate doubles

Second order

• Written as [A] ^2
• As you increase a you increase the rate^2
• e.g.
• 2[A] ^2= rate x4
• 3[A] ^2 = rate x 9
2 of 29

## the rate equation and rate constant

The rate equation is written as rate=k*[A] ^n [B] ^m (A and B represent the chemicals n and m represent the orders of reaction)

Often you will be asked to work out the rate constant the symbol for the rate constant is lowercase k so to work this out you rearrange the equation

K= rate of reaction/ [A] ^n [B] ^m

k will generally make four rate constants

• Zero order
• First order
• Second order
• Third order

To work out the units you

Units of rate/ [A]units ^n and [B] units ^m

E.g. question

• Work out the rate constant when A= 2nd order and B= first order
• Rate=k*[A] ^2[B] ^1
• k=Rate/[A] ^2[B] ^1
• the units
• k=Mol DM^-3 s/(Mols DM ^-3) ^2(Mol DM ^-3)= mols^-2 DM^6 s^-1

3 of 29

## overall order of reaction and half lives

The overal order of reaction is the sum of all the orders within a reaction

Half lives

• You can use concentration rate graphs to work out orders of different reactants
• Working out the half-lives are important features of these
• They can be used to work out the order of a reactant
• The definition for halve live the time taken for the concentration of the reactant to go down by half

Relating orders and half lives

Zero order

• On a time contrition graph a zero order reactant concentration decrees at a constant rate
• The half-life decrees with time

First order

• On a time contrition graph concentration half’s in equal time intervals
• Half-life is constant

Second order

• On a time contrition graph the concentration decrees rapidly in at first  but the slows down
• Half-life increase with time
4 of 29

## graphs for rate/concentration vs concentration/tim 5 of 29

## what order from rate/concentration graphs show

Zero order

increseing the constartion has no effeect on the rate of reaction

first order

the rate increse by the same amount as the reactant

if you doubel the conctartion of the reactant you doubel the rate

second order

as you incree the concentation of the reactant you ^2 the rate

e.g tripel reactant you increse the rate by 9 times

qadupel the reactant you increse the rate by 16 times

6 of 29

## how to determining the rate determining step

This can be easily worked from the rate equation

For example

Work out the rate determining step for this reaction

• NO2+CO====> NO + CO2
• NO2 is a second order and CO is a zero order
• As such the rate equation will be
• Rate=k [NO2] ^2 (since zero orders have no effect on the rate they can be removed)
• So the rate determining step has to involve NO
• And there has to be two NO as it is a second order
• So the rate determine step will be
• NO2+NO2=====>NO3+ NO (YOU MUST CHOSE SUTABEL PRODUCTS THESE ARE SUTIVBEL BEACUSE YOU ALREADY HAVE ONE PRODUCT)
• Step 2
• NO3+CO=====>CO2+NO2 (the NO2 acted as a cat lists and was regenerated)
• Overall equation
• NO2+NO2====>NO3+NO
• NO3+CO=====>NO2+CO2
• NO2+CO=====>NO+CO2
7 of 29

## equalibrum constant equation

the equilibrium constant Kc is the equilibrium constant and is put basically as Kc=product/reactant

the powers of the concentrations are the moles of the reactant in the equation

8 of 29

## equalibrum information

as an equilibrium reaction goes on you have a constant shift between products and reactants eventually it will reach a place where the reaction is in dynamic equilibrium

at this point the rate of the forward reaction is the same as the rate of the backward reaction

the concentrations of the reactants and products remain the same

the equilibrium constant has three main units mols DM^-3, DM^3 mols^-1 and can have no units as long as you show it has no units this is correct

equilibrium position and Kc

the magnitude of Kc indicates the extent of the chemical reaction

when Kc is more than 1 this means that the reaction is in favor of the products

when Kc is less than one the reaction is product favored

temperature and equilibrium

as you change the temperature you can have two effects

if you increase a temperature you shift the reaction in the endothermic direction

if you decrease the temperature you shift equilibrium in the exothermic direction

shifts in equilibrium are controlled by Kc its value only changes with temperature

9 of 29

## equalibrum constant vs the rate constant

changes in concentration

k = changes the rate unless you change the concentration of a zero order reactant. will cause an increase in the rate of reaction

Kc= if you increase the amount of a reactant it will shift back to the proper ratio so the value will stay the same

changes in pressure

K= this will cause the rate of reaction to increase as the reactants are occupying a smaller space so will have more successful collisions

Kc= again will have no effect as it will just shift it back to equilibrium

catalysts

k= it will cause the rate to increase as it will allow a lower activation energy for the reactants increasing the rate and, therefore, the rate constant

Kc=the catalysts speed up both the rate of the forward and backwards reaction so has no effect on the rate of equilibrium

10 of 29

## summary of rate and equalibrum constants

The Equilibrium constant

• the symbol for it is Kc
• it compares the concentrations and products and reactants present at equilibrium
• a large Kc value means that the reaction is product favoured and equilibrium is to the right
• a small Kc value means the reaction is reactant favoured and equilibrium is to the left
• if the forward reaction is endothermic the value of Kc will increase temperature increases
• if the forward reaction is exothermic the value of Kc will decrease as temperature increase
• you can write Kc from a balanced chemical equation

The Rate constant

• the symbol for it is k
• it measures the rate at which a specific reaction occurs
• a large value of k means a fast rate of reaction
• k increases with an increase in temperature
• it can only be worked out experimentally reactions often take several steps to reach the required products the rate is a snapshot of the slowest rate determining step
11 of 29

## Defining an Acid and base

Arrhenius acid

• this is the most basic definition we have for acids and bases
• it was based on the idea that some substances dissociated in solution to make a positive and negative ion
• from this he designed  an equation that is used to describe acids bases and neutralization equation is
• H^(+) + OH^(-) ====> H2O
• in this equation an acid is a H^+ hydrogen ion
• a base is an OH^- ion

Bronsted Lowery acid and base

• An acid is a proton donator
• A base is a proton acceptor
• this means that anything that can accept a proton is a base rather than just OH^- ion
• an example of a Bronsted Lowery acid-base reaction
• HCL + NH3=====> NH4^(+) + CL^(-)
12 of 29

## lewis acids and bases

lewis acid and base

• an acid is an electron acceptor
• A Lewis base is an electron donor
• 13 of 29

## the role of protons in acid reactions

when using acids you are mostly looking at the movement fo hydrogen ions or protons

depending on the formula of each acid and the strength of their bonding each acid can release aa different amount of protons
Theirs are three main types of basic acid these are acids that can comply dissociate their protons

the three types are mono, di and tri basic acids
a mono basic acid releases one hydrogen proton

• an example is HCl reaction with water
• HCl + H2O =====>H3O^(+)+Cl^-

a di basic acid can release two hydrogen ions

• an example is H2SO4 reacting with 2 mols of H20
• H2SO4 + H20====>H3O^(+) + HSO4^(-)
• HSO4^(-)+H2O====>H3O^(+)+SO4^(2-)

A Tri basic can release three hydrogen ions

• an example of is H3PO4 reacting with 3 mols of H2O
• H3PO4 + H20=====>H2PO4^(-) +H30^(+)
• H2PO4^(-) + H20==>HPO4^(2-)  + H30^(+)
• HPO4^(2-) + H20==>  PO4^(3-)  + H30^(+)
14 of 29

## the role of protons in acid reactions continued

Theirs are three main types of acid-base reactions

• acid and carbonate reactions
• acid and base reactions
• acid and alkali reactions

rules for ionic equations

• to Write an ionic equation for these reactions you need to cancel all species that do not change this includes physical state
• ions dissociate only in solution so a solid is usually shown undissociated

reactions with carbonates

• full equation= 2HCl(aq) + CaCO3(s)===>CaCL2(aq)+ CO2(g) + H20(l)
• ionic equation= 2H^(+) + CaCO3=====>Ca^(+) + H20 + CO2

reactions with bases

• full equation=2HNO3(aq) + MgO(s)====>Mg(NO3)2(aq) + H2O(l)
• ionic equation=2H^(+) + MgO  ======>Mg^(2+) + H20

reactions with alkilis

• full equation= HCl(aq) + KOH(aq) =====> KCl(aq)+H20(l)
• ionic equation=H^(+) + OH^(-)   =====> H2O
15 of 29

## the role of protons in acid reactions continued

Theirs are three main types of acid-base reactions

• acid and carbonate reactions
• acid and base reactions
• acid and alkali reactions

rules for ionic equations

• to Write an ionic equation for these reactions you need to cancel all species that do not change this includes physical state
• ions dissociate only in solution so a solid is usually shown undissociated

reactions with carbonates

• full equation= 2HCl(aq) + CaCO3(s)===>CaCL2(aq)+ CO2(g) + H20(l)
• ionic equation= 2H^(+) + CaCO3=====>Ca^(+) + H20 + CO2

reactions with bases

• full equation=2HNO3(aq) + MgO(s)====>Mg(NO3)2(aq) + H2O(l)
• ionic equation=2H^(+) + MgO  ======>Mg^(2+) + H20

reactions with alkilis

• full equation= HCl(aq) + KOH(aq) =====> KCl(aq)+H20(l)
• ionic equation=H^(+) + OH^(-)   =====> H2O
16 of 29

## Redox reactions of acids with metals

the reaction of an acid with a metal is a redox reaction so doesn't count as an acid-base reaction

• acids react with an acid to form salt and hydrogen gas
• full equation=2HCl(aq)+Mg(s)====> MgCl2(aq) + H2
• ionic equation= 2H^(+) + Mg (s)==> Mg^(2+)   + H2
17 of 29

## congigate acid and base pairs since most acid-base reactions are reversible. when acids react with a base in these situations the base accepts a proton and the acid donates a proton. however, the base has now become an acid as it can now donate the proton it has just gained and the acid is now a base as it can now accept a proton. this is how conjugate acid-base pairings work.

18 of 29

## what is pH

• pH is defined as the -base log of the concentration of hydrogen ions in a solution
• pH is measured on a scale that starts at -1 all the way up to pH14

what does a pH value tell us

• a low pH= high concentration of H^(+)
• as such it will be very acidic
• a high pH= low concentration of H^(+)
• a pH change of one changes the value for the concentration of H^(+) by a 1000 times

the amount of hydrogen ions can be worked out by base ten 10 - the pH value

19 of 29

## strong acids

a strong acid is an acid that completely dissociates in solution 20 of 29

## weak acids

a weak acid is an acid that only partially dissociated in solution as you can see the acid only releases some protons in a weak acid

21 of 29

## calculating pH for strong acids

strong acids undergo complete dissociation in aqueous solution

in order to work out the pH you need the acid dissociation constant

the acid dissociation constant equation is written as such

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

however this can be simplified for strong acids as they completely dissociate you can simplify the equation to

[H^+]=[acid]

e.g

1.22x10^-3 mol HCl completely disassociates work out the pH for the acid

[H^+]=[HCl]

so the H^+ ion concentration is 1.22x10^-3

pH=-log[1.22x10^-3] gives a pH of 2.91

this can be used for all strong acids

22 of 29

## working out pH for weak acids

to calculate the pH for weak acids we use the acid dissociation constant and change it slightly

the acid dissociation constant for a weak acid = Ka=[H^+]^2/[HA]

you than rearrange the equation to make [H^+]^2= Ka[HA]

then square root both sides to make [H^+]=Sr of Ka[HA]

e.g

you have HNO2 with a concentration of is 0.055mols the acid dissociation constant for this is Ka=4.70x 10^-4

plug this into the equation and you get

[H^+]=sqare root of 4.70x10^-4[0.055]= 5.08x10^-3

-log[5.08x10^-3]= 2.29

23 of 29

## the ionization of water & the water dissociation c

the water dissociation constant at 25^oC is Kw= 1.00x10^-14

why is Kw so important?

this controls the balance between [H^+] and [OH^-] in all aqueous solutions.

the relative concentrations of acidity and basicity are measured against the Kw.

the reason, why water has a pH of seven, is that together both the [H^+] and [OH^-] must add up to make -14

so if in water [H^+]=[OH^-] so each must have a concentration of

1.00x10^-7

the -log [1.00x10^-7]= 7

24 of 29

## values of bases and working out POH

the strength of a base is measured by its ability to produce OH^- ions in solution

some examples of this include NaOH and NH3

NaOH(aq) =====> Na^(+)+OH^- this is a strong base as it completly dissoctates OH^- ions in soultion

NH3 + H2O<====> NH4^(+)+ OH^(-) this is a weak base as it only partially dissociates OH^- ions and the position of equilibrium favours the left

pOH is the opposite of pH it works out how basic something is

to work out pOH, you get the -log of the[OH^-] you then subtract this number from 14 this will give you your pOh value.

and to work out the concentration of OH you use this equation for strong bases

[OH^-]=[base]

e.g

0.0050 mols dm^-3 of KOH at 25^oC or 298 K

pOH=-log[0.0050]

pOH= 14-2.30= 11.7

25 of 29

## buffer solutions

• these are solutions that resist changes of pH when small amounts of acid or alkali are added
• a buffer solution is a mixture of a weak acid and its conjugate base

how does it act?

• the weak acid and the conjugate base are both responsible for this ability to resist changes in pH upon the addition of an acid or base

the overall principle is simple the conjugate base reacts with the acid removing it

like so

[H^+] incresed
the conjugate base reacts with it, equilibrium shifts to the left removing most of the acid

•       equlibrum
• <================
• HA<======> H^(+)+A^-

the weak acid reacts with the [OH^-] ions

• the acid releases more H^+ to restore the balance
• equilibrium shifts to the right
• equilibrium
• ================>
• HA<======> H^(+)+A^-
26 of 29

## how to work out the ph value of a buffer solution

to work out the pH of a buffer solution you use this equation

[H^+]= Ka x [weak acid]/[congigate base]= answer

e.g a buffer solution contains 0.15 mols dm^-3 of methanoic acid and 0.065 mols dm^-3 of sodium methanolate its Ka =1.6 x10^-4

calculate the pH

[H^+]=1.6 x10^-4 x [0.15]/[0.065]= 0.000369

-log[0.000369]= 3.43

27 of 29

## Indicators

There are three main indicators that we need to know

phenolphthalein

this turns from pink to clear and colourless

it is good for reactions that have a neutral point of about 7

bromomethyl blue

changes from blue to yellow

green at its neutral point

better for a more basic neutral point

methyl orange

goes from a red/pink to yellow

orange at neutral point

better at more acidic neutral points

28 of 29

## standard enthalpy change of neutralisation

to calculate the energy change of neutralisation you first need to work out the energy change of the solution

you do this with the specific heat capacity equation #

energy change = mass x specific heat capacity x change in temperature

work out the moles that reacted

using n=CxV/1000

scale all the moles in a chemical equation

so if it is a one: one ratio both would have the same amount of moles yu worked out

divide the energy change by the moles

if all the ionic equations to make water for a neutralisation reaction are similar the enthalpy changes for neutralisation will be similar

e.g

HBr + NaOH ======> NaBr + H2O = enthalpy change = -57.6

H^(+)+OH^-=====> H2O

HNO3+NaOH=====>H2O + NaNO3 =  enthalpy change = -57.6

H^(+)+OH^-=====> H2O

29 of 29