# Chemistry Unit 4- Kinetics and Equilibria

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• Created on: 10-06-11 09:26

## Rate Equations

The reaction rate tells you how fast reactants are converted to products

If reactants are in solution, the rate will be the change in concentration per second, and the units will be moldm-3 s-1

Rate Equation- tells you how the rate is affected by the concentrations of reactants.

General reaction = A + B --> C + D

Rate Equation = K [A] ^m [B] ^ n               with units of rate as moldm-3s-1

m and n are the orders of the reaction with respect to reactants A and B .m tells you how the concentration of reactant A affects the rate and n tells you the same for reactant B

If [A] changes and the rate stays the same, the order of the reaction with respect to A is 0. (i.e. if [A] then doubles or triples, the rate stays the same). If the rate is proportional to [A], the order of reaction with respect to A is 1. (i.e. if[A] doubles, the rate will double. if [A] triples, the rate will triple). If the rate is proportional to [A]^2, then the order with respect to A is 2. (i.e. if [A] doubles, the rate will be 4 times faster. if [A] triples, the rate will be 9 times faster).

The overall order of the reaction is m + n

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## K is the rate constant

The bigger the rate constant, the faster the reaction.

The rate constant is always the same for a certain reaction at a particular temperature.

When you increase the temperature, the rate constant rises:

It increases the number of collisions between reactant molecules, and so the energy of each collision. But the concentrations of the reactants and orders of the reaction stay the same. So the value of K must increse for the rate equation to balance.

FYI---> [X]^1 is usually written as [X]

[X] ^0 equals 1 so is usually left out of the rate equation

(Even though some things are catalysts rather than reactants, they can still be in the rate equation).

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## Initial Rates Method

Initial rate of a reaction is the rate at the start of a reaction

Find it from concentration-time graph by calculating the gradient of the tangent at time= 0

How to use the method:

1. Repeat experiment several times using different initial concentrations of reactants. You should usually only change one concentration at a time, but they might be mean in the exam.

2. Calculate the initial rate for each experiment using above method

3. See how the initial concentrations affect the initial rates and figure out the order of each reactant.

Pretty much in the exam, they like to use that table thing and work it out using experiments 1 and 3 then 2 and 3 etc.... just don't panic :)

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## Rate determining step

Mechanisms can have 1 step or a series of steps. Each step can have a different rate. The overall rate is decided by the step with the slowest rate-->the rate-determining step (otherwise known as the rate-limiting step).

If a reactant appears in the rate equation, it must affect the rate. So this reactant, or something derived from it must be in the rate-determining step. If the reactant doesn't appear, it won't be in the determining step, nor anything derived from it.

Catalysts can also be rate-determining steps.

1. The rate-determining step doesn't have to be the first step in a mechanism

2. The reaction mechanism can't usually be predicted from just the chemical equation

The order of a reaction with respect to a reactant shows the number of molecules of that reactant that are involved in the rate-determining step. e.g. if  reactions second order with respect to x, they'll be two molecules of x in the rate determining step.

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## Rate-determing step and reaction mechanisms

E.g. There are two possible mechanisms for the nucleophile OH- substituting for Br in 2-bromo-2-methylpropane:

1. CH3CH3CCH3Br + OH-  --> CH3CH3CCH3OH + Br-      or:

2. CH3CH3CCH3Br --> CH3CH3C+CH3 + Br-  as step one and then:

2. CH3CH3C+CH3 + OH- --> CH3CH3CCH3OH as the second

The rate equation was worked out using rate experiments: Rate= K[(CH3)3 CBr]

OH- isn't in the rate equation so it can't be involved in the rate determining step. Second mechanism is correct as OH- isn't in the rate-determining step.

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## Rate-determining step and reaction mechanisms

E.g.2. = Nitrogen monoxide can react with oxygen to produce nitrogen dioxide:

2NO + O2 --> 2NO2   and rate = K [NO]^2 [O2]

The rate determining step must involve 2 molecules of NO and 1 of O2

Reaction mechanism is made up of two steps:

1. NO + NO --> N2O2

2. N2O2 +O2 --> 2NO2

Neither step 1 or step 2 contains the molecules you'd expect from rate reaction.

But in step 2 there's an intermediate molecule - N2O2 that is derived from two molecules of NO.

So step two is the rate-determining step.

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## Calculating Rate Constant from the Orders and Rate

(before starting remember units of K vary)

E.g. Reaction with second order with respect to No and zero order with respect to CO and O2. At certain temperature, the rate is 1.76 x 10^-3 moldm^-3s^-1 when [NO(g)]= [CO(g)] = [O2 (g)] = 2.00 x 10^-3 moldm^-3. Find the value of the rate constant, K , at this temperature.

NO(g) + CO(g) + O2(g) --> NO2(g) + CO2(g)

1. Write out the equation: Rate = K[NO]^2[CO]^0[O2]^0 = K[NO]^2

2. Insert concentration and rate. Rearrange and calculate value of K:

Rate = K[NO]^2 so...

1.76 x 10^-3 = K x (2.00 x 10^-3)^2   --->  K = 1.76 x 10^-3/ 2.00 x 10^-3 = 440

Find units of K = moldm-3s-1/ (moldm-3)^2 = s-1/moldm-3 = dm^3mol^-1s^-1

so the answer is K= 440 dm^3mol^-1s^-1

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## The equilibrium constant

Lots of changes are reversible

As reactants get used up, the forward reaction slows down- and as more product formed, the reverse reaction speeds up.

After a bit, the forward reaction will go as the same rate as the backward reaction.The amounts of reactants and products wont be changing anymore, so it'll seem like nothings happening- called dynamic equilibrium

Dynamic equilibrium can only happen in a closed system at a closed temperature.

Kc is the equilibrium constant

If you know the molar concentration of each substance at equilibrium, you can work out Kc, but it will only be true for that particular temperature.

General reaction aA + bB <--> dD + eE (lower case letters are the no.of moles)

Kc = [D]^d x [E] ^e / [A] ^a x [B]^b

Remember - the little numbers here aren't the rate order, they are the number of moles.

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## Position of Equilibrium

If conditions change, the position of equilibrium will move:

Changes can be concentration, pressure or temperature. Altering the position will mean you'll end up with different amount of reactants and products at equilibrium

If the position of equilibrium moves to the left, you'll get more reactants:

H2 (g) + I2 (g) <--> 2HI (g)

If the position of equilibrium moves to the right, you'll get more products:

H2 (g) + I2 (g) <--> 2HI (g)

Le Chatelier's Principle:

If there's a change in concentration, pressure or temperature the equilibrium will move to counteract the change

Pretty much, raise the temp and the equilibrium will shift to try to cool it, if you raise pressure or concentration, the equilibrium will shift to reduce it.

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## What alters Kc:

Temperature:

Increasing temperature adds heat. Equilibrium shifts in endothermic direction to absorb the heat

Decreasing temperature removes heat. Equilibrium moves in exothermic direction to replace heat

If forward is exothermic then backwards is endothermic and vice versa. If the change means more product is formed (shift to the right) then Kc will rise. If the change means less is formed (shift to the left) then Kc will decrease

Concentration:

The value of the equilibrium constant, Kc, is fixed at any given temperature. So if the concentration of one thing in the equilibrium mixture changes, then the concentrations of the others must change to keep Kc the same

Catalysts:

Have no effect on equilibrium position or value of Kc. Can't increase yield but do mean equilibrium is reached faster

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## Acids and Bases

Bronsted-Lowry acids= Proton donors, release H+ ions when mixed with water. (but H+ always combined with H2O to form hydroxonium ions, H3O+):

HA (aq) + H2O (l) --> H3O+ (aq) + A-(aq)      HA is just any acid

Bronsted-Lowry Bases= Proton acceptors. When in solution, they grab H+ ions from water molecules:

B (aq) + H2O (l) --> BH+(aq) + OH- (aq)    B is just any base

Strong acids fully dissociate in water - nearly all H+ions are released. Hydrochloric acid is a strong acid:  HCl (g) + water --> H+ (aq) + Cl- (aq)

Strong bases e.g. sodium hydroxide ionise almost completely in water: NaOH (s) + water --> Na+ (aq)  OH- (aq) . They are reversible, but equilibrium lies to extremely far to the right

Weak acids e.g. ethanoic/ citric dissociate slightly, and the equilibrium lies well to the left: CH3COOH (aq) <--> CH3COO-(aq) + H+ (aq)

Weak bases e.g. ammonia only slightly dissociate, again equilibrium well left: NH3 (aq) + H2O(l) <--> NH4+(aq) + OH- (aq)

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## Water...

Acids can only get rid of their protons if there is a base to accept them:

HA (aq) + B (aq) <--> BH+ (aq) + A- (aq)

It's an equilibrium, so if you add more HA or B, position will move to the right. But if you add BH+ or A-, the equilibrium will move left

When an acid is added to water, water acts as a base and accepts the proton (equilibrium is far too right for strong acids, and too left for weak acids):

HA (aq)  H2O (l) <--> H3O+ (aq) + A- (aq)

Water dissociates into hydroxonium ions and hydroxide ions:

H2O (l) + H2O (l) <--> H3O+(aq) + OH- (aq)  or simply:

H20 (l) <--> H+(aq) + OH- (aq)

So the equilibrium constant for water is:

Kc = [H+] x [OH-] / [H2O]

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## Kw

Water only dissociates a tiny amount, so equilibrium lies well left.

So much water compared to H+ and OH- the concentration of water is considered to have a constant value

If you multiply Kc for water ( constant) by [H2O] (another constant) you get a constant.

New constant is called the ionic product of water - Kw

Kw = Kc x [H2O] = [H+] [OH-] .... Kw = [H+] x [OH]  units are always mol^2dm^-6

Kw always has the same value for an aqueus solution at a given temperature.

In pure water, there's one H+ for one OH- so [H+] = [OH-] so if dealing with pure water, you can say that Kw = [H+] ^2

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## pH

Conc. of H+ ions in solution varies enormously, so we use a logarithmic scale:

pH= -log10 [H+]        where [H+] is measures in moldm^-3

pH scale goes from 0 (very acidic) to 14 (very alkaline). pH 7 is seen as neutral

1. Can calculate pH from H+ concentration:

e.g. concentration of ions in hydrochloric acid is 0.01 moldm^-3

pH= -log10 [0.01] = 2

2. Can calculate H+ ion concentration from pH: use [H+] = 10 ^-pH

e.g. sulfuric acid has a pH of 1.52

[H+] = 10 ^-1.52 =0.03 moldm^-3 .... = 3 x 10^-2 moldm^-3

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## Monoprotic and Diprotic:

Strong acids like hydrochloric and nitric acid (HNO3) ionise fully

They are monoprotic- one mole of acid will release one proton when it dissociates each, so mole of acid produces one mole of hydrogen ions.

This means the H+ concentration is the same as the acid concentration:

0.1 moldm^-3 HCl has [H+] = 0.1 moldm^-3 so pH= -log10 o.1 = 1.0

Each mole of diprotic acid releases two protons when it dissociates.

E.g. sulfuric acid: H2SO4 (l) + water --> 2H+ (aq) + SO4 2- (aq)

So diprotic acids produce two moles of hydrogen for each mole of acid.

This means that the H+ concentration is twice the concentration of the acid:

0.1 moldm^-3 H2SO4 has [H+] 0.2 moldm^-3 so pH = -log10 0.2 = 0.70

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## Use Kw to find pH of a base:

Sodium hydroxide (NaOH) and Potassium hydroxide (KOH) are strong bases.

They fully ionise in water and donate 1 mole OH- ions per mole of base

Means the concentration of OH- ions is the same as the concentration of the base

so for 0.02 moldm^-3 NaOH, [OH-] is 0.02 moldm^-3

To work out pH you need to know [H+] , but use Kw to help:

Kw = [H+] [OH-] ..... rearrange to [H+] = Kw/ [OH-]

e.g. value of Kw at 298K is 1.0 x 10^-14 mol^2dm^-6. Find pH of 0.1 moldm^-3 NaOH at 298 K.

[OH-] = 0.1 moldm^-3  ----->   [H+] = 1.0 x 10^-14 / 0.1 = 1.0 x 10^13 moldm^-3

so pH = -log10 1.0 x 10^-13 = 13.0

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## Use Ka to find pH of a weak acid

Weak acids don't ionise fully, so [H+] isn't the same as acid concentration

For weak aqueous acid, HA, you get the equilibrium HA(aq) <--> H+(aq) +A-(aq)

Only a tiny amount of HA dissociates, so assume that [HA] start = [HA] equilibrium

Apply the law: Ka = [H+] x [A-] / [HA]

You can assume all the H+ ions come from the acid so [H+] = [A-]

So, Ka = [H+]^2 / [HA]    and the units of Ka are moldm^-3

e.g. Find pH of a 0.02 moldm^-3 of propanoic acid at 298K. the Ka at this temperature is 1.30 x 10^-5 moldm^-3.

Find expression for Ka, then rearrange to find [H+]^2:

Ka= [H+]^2 / [CH3CH2COOH] --> [H+]^2 = Ka [CH3CH2COOH]

(1.30 x 10^-5) x 0.02 = 2.60 x 10^-7. so then [H+] = √2.60 x 10^-7

=5.10 x 10^-4 moldm^-3 so then ... pH= -log10 5.10 x 10^-4 = 3.29

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## FInd concentration or Ka of a weak acid

E.g. pH of ethanoic acid (CH3COOH) sol is 3.02 at 298K. Calculate conc. of this solution. The Ka of acid is 1.75 x 10^-5 moldm^-3 at 298K.

Use pH to find [H+] ---> [H+] = 10^-pH = 10^-3.02 = 9.55 x 10^-4 moldm^-3

Then write expression for Ka and rearrange to equal [CH3COOH]:

Ka = [H+]^2 / [CH3COOH] --> [CH3COOH] = [H+]^2/ Ka

= (9.55 x 10^-4)/ 1.75 x 10^-5 = 0.0521 moldm^-3

E.g. sol of 0.162 moldm^-3 HCN has a pH of 5.05 at 298K. Find value for HCN:

Use pH of acid to find [H+]:

[H+] = 10^-pH = 10^-5.05 = 8.91 x 10^-6 moldm^-3

Write expression for Ka and substitute in values for [H+] and HCN:

Ka= [H+]^2/ [HCN] = (8.91 x 10^-6)^2/ 0.162 = 4.90 x 10^-10 moldm^-3

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## pKa

pKa is calculated from Ka in the same way as pH is calculated from [H+]

pKa = -log10Ka ...........  Ka= 10^-pKa

So if acid has a Ka value of 1.50 x 10^-7 then pKa = -log10 (1.50 x 10^-7)= 6.82

If an acid has a pKa value of 4.32, then Ka = 10^-4.32 = 4.79 x 10^-5

May have a pKa value in "find the pH" type of question. Convert it to Ka, so you can use the Ka expression:

e.g. Calc. the pH of 0.050 moldm^-3 methanoic acid (HCOOH), and has a pKa of 3.75 at 298K.

Ka= 10^-pKa = 10^-3.75 = 1.78 x 10^-4 moldm^-3 (have to convert pKa to Ka)

Ka= [H+]^2 / [HCOOH] -> [H+]^2 = Ka [HCOOH]

= 1.78 x 10^-4 x 0.050 =8.9 x 10^-6

= [H+] = √8.9 x 10^-6 = 2.98 x 10^-3 moldm^-3  pH= - log 2.98 x 10^-3 = 2.53

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## Titrations

Titrations allow you to find exactly how much alkali is needed to neutralise a quantity of acid.

1. Measure out acid of known concentration using a pipette and put in flask, along with the appropriate indicator.

2. Do a rough titration, add alkali to acid using a burette, swirl glass until you find the approximate end point

3. Do an accurate titration. Record the amount of alkali needed to neutralise the acid. Repeat to get concordant results (within 0.1cm3 of each other).

Indicators:

For strong acid/ strong alkali use methyl orange (yellow at high pH and red at low) or phenolphthalein (pink at high pH and colourless at low pH).

For strong acid/ weak alkali use methyl orange, range is 3.1-4.4

For weak acid/ strong alkali use phenolphthalein, range is 8.3-10

For weak acid/ weak alkali there's no sharp pH change so no indicator will work.

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## Calculating titrations...

Pretty simple steps:

1. Write a balanced equation

2. Work out the moles of known

3. Work out the ratio

4. Work out moles of the unknown

5. Work out concentration

Things to remember:

-Divide by 1000 to get volume from cm3 to dm3

-Moles = concentration x volume (cm3)/ 1000

-Concentration = moles x 1000/ volume (cm3)

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## Equivalence point

You can use a pH curve to work out how much acid or base needed if you have used a pH meter rather than an indicator.

Find the equivalence point- the mid-point of the line of rapid pH change and draw a vertical line downwards until it meets the x-axis.

The value at this point on the x-axis is the volume of acid or base needed.

A diprotic acid can release two protons. e.g. ethanoic acid. When it reacts with a base like sodium hydroxide it's neutralised

Reaction happens in two stages, becasue the two protons removed seperately

The pH curve will have two equivalence points. e.g. at pH  2.7 and 8.4

You can calculate the concentration of a diprotic acid from titration data in the same way as you did for a monoprotic acid.

However, because it's a diprotic acid, you need twice as many moles of base as moles of acid

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## Buffer Action

Buffer= a solution that resists changes in pH when small amounts of acid or alkali are added.

You get acidic buffers and base buffers.

Acidic Buffers:

Have a pH of less than 7, made by mixing a weak acid with one of it's salts.

e.g. ethanoic acid (CH3COOH) and sodium ethanoate (CH3COO-Na+)

Ethanoic acid is weak so only slightly dissociates:

CH3COOH (aq) <--> H+ (aq) + CH3COO- (aq)

But the salt fully dissociates into its ions when it dissolves:

CH3COONa (s) + water --> CH3COO- (aq) + Na+ (aq)

In the solution you've got undissociated ethanoic acid molecules and ethanoate ions.

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## Acidic Buffers

When you alter the concentration of H+ or OH- ions in the buffer solution the equilibrium position moves to counteract the change- remember Le Chatelier's.

CH3COOH (aq) <--> H+ + CH3COO- (aq)

(undissociated acid)            (from salt)

Add a small amount of acidH+ conc. increases. Most of the extra H+ ions combine with the CH3COO- ions to form CH3COOH, shifting the equilibrium to the left, reducing the H+ concentrations to close to it's original value. So pH doesn't change. (Large number of CH3COO- ions make sure that the buffer can cope).

Add a small amount of base: (e.g. NaOH), the OH- concentration increases. Most of the extra OH- ions react with H+ ions to form water, removing H+ from the solution. Causes more CH3COOH to dissociate to form H+ ions (no problem as there's loads spare). Equilibrium shifts to the right. The H+ conc. increases until it's close to its original value so the pH doesn't change

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## Basic Buffers

Basic buffers have  pH greater than 7, and are made by mixing a weak base with on of its salts

E.g. Ammonia (NH3, weak base) + ammonium chloride (NH4Cl, salt of ammonia)

Salt fully dissolves in solution: NH4Cl (aq) + NH4+(aq) + Cl-(aq)

Some NH3 molecules will also react with water molecules:

NH3 (aq) + H2O (aq) <--> NH4+(aq) + OH- (aq)

So the solution will contain lots of ammonium ions (NH4-) and ammonia too.

(Buffers are used in shampoo, a pH 5.5. to counteract the alkaline soap. Biological washing powders contain buffers to keep pH level right for enzymes and we have biological buffer systems in our bodies, making sure tissues stay at the right pH. Our blood needs to stay near a pH close to 7.4).

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## Basic Buffers continued...

the equilibrium position of this reaction can move to counteract changes in pH:

NH3 (aq) ((lots of)) + H2O (l) <---> NH4+ (aq) ((lots of )) + OH- (aq)

Add a small amount of base: OH- concentration increases, making the solution more alkaline. Most of the extra OH- ions will react with NH4+ ions, to form NH3 and H2O. So the equilibrium will shift to the left, removing OH- ions from the solution, and stopping the pH from changing much.

Add a small amount of acid: H+ concentration increases, making the solution more acidic. Some of the H+ ions react with OH- ions to make H2O. When this happens the equilibrium position moves to the right to replace the OH- that have been used up. Some of the H+ ions react with NH3 molecules to form NH4+ ---> NH3 + H+ <--> NH4+. These reactions will remove most of the extra H+ ions that were added, so the pH won't change much

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## Calculating pH of Buffer Solution

To calculate pH of an acidic buffer you need to know the Ka of the weak acid and the concentrations of the weak acid and its salt:

e.g. buffer solution contains 0.40 moldm^-3 methanoic acid (HCOOH) and 0.6 moldm^-3 sodium methanoate (HCOO- Na+). For methanoic acid, Ka = 1.6 x 10^-4 moldm^-3. Calculate the pH:

1. Write the expression for Ka of the weak acid:

HCOOH (aq) <--> H+ (aq) + HCOO- (aq)

so Ka= [H+ (aq)] x [HCOO- (aq)] / [HCOOH (aq)]

2. Rearrange the expression and put in data to calculate [H+(aq)]:

[H+(aq)] = Ka x [HCOOH(aq)] / [HCOO- (aq)]

[H+(aq)] = 1.6 x 10^-4 x (0.4/0.6) = 1.07 x 10^-4moldm^-3

3. Convert [H+(aq)] to pH: pH = - log10 (1.07 x 10^-4) = 3.97

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## Assumptions that need to be made...

* HCOO-Na+ is fully dissociated, so assume that the equilibrium concentration of HCOO- is the same as the initial concentration of HCOO-Na+ *

* HCOOH is only slightly dissociated, so assume that its equilibrium concentration is the same as its initial concentration *

* And we are done :) *

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