## Tuesday, 26 May 2015

### Building and balancing chemical equations (2) Combustion of hydrocarbons

How can you build combustion equations for both complete and incomplete combustion of hydrocarbons?

I know there are many of you out there who are struggling with how to build hydrocarbon combustion equations.

I think my approach is pretty fool proof so let me know what you think.

Can you construct complete combustion equations for hydrocarbon fuels?

Say you are asked to construct the equation for the complete combustion of propane: how do you do that?  (Thankfully they give you the formula of propane C3H8!!!!)

Here is a routine which if you memorise it you can’t fail to get these equations right

1  Complete combustion means just two products water and carbon dioxide so we can write the formulas of reactants and products like this:

C3H8   +      O2        =          CO2      +   H2O

2  Next you can see there are three carbon atoms in propane so that gives three CO2 molecules so the equation is now:

C3H8   +      O2        =          3CO2      +   H2O

3  Next you can see there are eight hydrogen atoms in propane but as two appear in water we'll divide eight by two (difficult!!!) and have four water molecules on the right like this:

C3H8   +      O2        =          3CO2      +   4H2O

4  All that's left is to find how many oxygens there are on the right.  I can count ten atoms so divide by two as two atoms make an oxygen molecule and that gives us five O2 on the left:

C3H8   +      5O2        =          3CO2      +   4H2O

5  The final equation then becomes:

C3H8   +      5O2        =          3CO2      +   4H2O

This method always works for these hydrocarbon combustion equations.

The only adjustment you'll have to make is if the hydrocarbon has an even number of carbons (e.g. butane C4H10) in it.

Then you will have to double every final value for each reactant and product to be left with a whole number of oxygen molecules.

Try it and see what I mean.

So having cracked the complete combustion equations how do you feel about the incomplete combustion equations?

Say you are asked to construct the equation for the incomplete combustion of pentane (C5H12) into water and carbon monoxide, carbon dioxide and carbon!

1  Let's lay the symbols down first:

C5H12   +  O2     =    C        +       CO       +        CO2     +    H2O

2  Now things get really interesting because there is not one straight answer.

You have five carbons in pentane and three products that all contain carbon.

Where do you put the carbons?

We could have three carbon atoms, one carbon monoxide and one carbon dioxide molecule.

Or we could have two carbon atoms, two carbon monoxide molecules and one carbon dioxide molecule!!

In other words, there are several possible equations so we have just to pick one.

So let's go with this:

C5H12   +  O2     =    3C        +       CO       +        CO2     +    H2O

3  Next there are twelve hydrogen atoms in pentane so dividing by two gives us six water molecules.  Like this:

C5H12   +  O2     =    3C        +       CO       +        CO2     +    6H2O

4  Lastly, we'll count up the oxygen atoms on the right (nine in all) and divide by two to find the number of oxygen molecules like this:

C5H12   +  O2     =    3C        +       CO       +        CO2     +    6H2O

5  But that's four and a half oxygen molecules which some of us might not be too happy with as it looks untidy (!!) so let's double up all the quantities of atoms and molecules and we'll all feel safer like this:

2C5H12   +  9O2     =    6C        +       2CO       +        2CO2     +    12H2O

Can you come up with some alternative equations for this same reaction?

### Building and balancing chemical equations (1) Neutralisation

How to build neutralisation equations

If combustion equations present us with problems to solve them so do neutralisation equations

Did you know there are at least two types of neutralisation equation?

Essentially neutralisation occurs when an acid neutralises a base or an alkali (an alkali being a water soluble base)

Here are three typical neutralisation equations:

First

Hydrochloric acid  +   sodium hydroxide   =         sodium chloride  +  water
HCl     +                     NaOH                      =         NaCl               +          H2O

This is an example of an acid neutralising a water soluble base (alkali)

Second

Nitric acid +   sodium carbonate = sodium nitrate + water + carbon dioxide
HNO3       +   Na2CO3                = NaNO3            +   H2O  +  CO2

This is an example of an acid plus an insoluble base i.e. a carbonate

Third

Sulfuric acid  +  magnesium oxide = magnesium sulphate + water
H2SO4          +       MgO                 =              MgSO4         +   H2O

This is an example of an acid plus an insoluble base i.e. a metal oxide

So how do we construct and balance these kinds of equations?

A) Let’s look at the first example: acid and soluble base

How can we build an equation for the reaction between say nitric acid and potassium hydroxide?

Let’s put the words down first:

This is the pattern:

Acid   +   soluble base   = water  +  soluble salt

Nitric acid  +  potassium hydroxide = ????

Next add the formulas of the two reactants so:

Nitric acid  +  potassium hydroxide = ?????
HNO3        +  KOH   =

Now notice that water forms so remove the elements of water from the reactants.

I have highlighted these elements so

Nitric acid  +  potassium hydroxide = ?????
HNO3        +  KOH   =

These will combine to form one water molecule

Nitric acid  +  potassium hydroxide = water
HNO3        +  KOH                          =   H2O

The remaining parts of the reactants form the soluble salt potassium nitrate

Nitric acid  +  potassium hydroxide = water  +  potassium nitrate
HNO3       +   KOH                          =   H2O   +   KNO3

With a diprotic acid like sulphuric acid things gets a little trickier

B) Let’s look at a second example: acid and insoluble base a metal oxide

How can we build an equation for the reaction between sulfuric acid and copper oxide?

Let’s put the words down first:

This is the pattern:

Acid   +   insoluble metal oxide   =  water  +  soluble salt

sulfuric acid  +  copper oxide = ????

Next add the formulas of the two reactants so:

sulfuric acid  +  copper oxide = ????
H2SO4           +   CuO

Now notice that water forms so remove the elements of water from the reactants.

I have highlighted these elements so

sulfuric acid  +  copper oxide = ????
H2SO4           +   CuO

These will combine to form one water molecule
sulfuric acid  +  copper oxide = water
H2SO4             +   CuO          =    H2O

The remaining parts of the reactants form the soluble salt copper sulfate

sulfuric acid  +  copper oxide = water   +   copper sulfate
H2SO4             +   CuO        =    H2O    +     CuSO4

With a monoprotic acid like nitric acid things gets a little trickier

C) Let’s look at a third example: acid and insoluble base a metal carbonate

How can we build an equation for the reaction between nitric acid and magnesium carbonate?

Let’s put the words down first:

This is the pattern:

Acid   +   insoluble metal carbonate   =  water  +  carbon dioxide + soluble salt

nitric acid  + magnesium carbonate  = ????

Next add the formulas of the two reactants so:

nitric acid  + magnesium carbonate  = ????
HNO3        +    MgCO3   =……….?

Now notice that water and carbon dioxide form so remove the elements of water and carbon dioxide from the reactants.

I have highlighted these elements so

nitric acid  + magnesium carbonate  = ????
HNO3        +    MgCO3   =……….?

These will combine to form one carbon dioxide molecule but there is not enough hydrogen for a water molecule, so we have to double up the nitric acid formulas.

nitric acid  + magnesium carbonate  = water   +  carbon dioxide
2HNO3      +    MgCO3                      =  H2O   +       CO2

The remaining parts of the reactants form the soluble salt magnesium nitrate and note that there are now two nitrates to one magnesium

Note too that the nitrates are placed in brackets with the 2 outside to show the magnesium to nitrate ratio is 1:2.

nitric acid + magnesium carbonate = water + carbon dioxide + magnesium carbonate
2HNO3     +    MgCO3                     =  H2O   +        CO2    +             Mg(NO3)2

You need to be able to make yourself familiar with as many examples like these as you can imagine for example:

Nitric acid   +  sodium hydroxide
Sulphuric acid  + potassium hydroxide
Hydrochloric acid  +  ammonium hydroxide
Sulphuric acid   +   ammonium hydroxide

Nitric acid  +   copper oxide
Nitric acid  +   zinc oxide
Nitric acid  +   iron (III) oxide

Sulfuric acid  +   magnesium oxide
Sulphuric acid   +   sodium oxide
Sulphuric acid   +   iron (II) oxide

Hydrochloric acid   +   copper oxide
Hydrochloric acid   +   iron (II) oxide
Hydrochloric acid   +   magnesium oxide

Nitric acid  +  copper carbonate
Sulphuric acid   +  potassium carbonate
Hydrochloric acid   +  ammonium carbonate  etc……….

## Monday, 25 May 2015

### Halogenoalkanes (7) Elimination reactions

Reactions of the Halogenoalkanes

How does competition between substitution and elimination reactions in haloalkanes work out?

What we find is this:

A) In ethanol solvent where OHacts as a base rather than a nucleophile then the corresponding alkene forms by elimination of HBr like this:

CH3CH2CH2CH2Br     +    OH       =      CH3CH2CH=CH2    +   H2O   +   Br
1-bromobutane                    base               but-1-ene

B) But if the solvent is changes for water then the substitution reaction dominates as the hydroxide ion acts as a nucleophile in the more polar solvent:

CH3CH2CH2CH2Br   +          OH   =          CH3CH2CH2CH2OH   +   Br
1-bromobutane                                            butan-1-ol

The important point to remember is that both substitution and elimination mechanisms occur together.

The use of a solvent favours one over the other but does not mean that one works exclusively instead of the other.

So ethanolic potassium hydroxide allows the elimination mechanism to predominate

Aqueous potassium hydroxide favours the substitution mechanism in the haloalkane.

I have discussed the substitution mechanism here

The elimination mechanism is a concerted mechanism, one step following on the next…

First, the base abstracts a proton attached to the carbon atom attached to the C—Br bond.

Second, the two electrons remaining from the C—H bond form a double carbon-carbon bond

Third, these two electrons repel the electrons in the C—Br bond (rather than the C—CH3 bond) because the bromine atom forms a good leaving group as a bromide ion (Br).

You ought also to realise that breaking a C—H or a C—CH3 bond costs much more energy than breaking a C—Br bond.

The result is an alkene see below:

In this example, 2-bromo-2-methyl propane is converted into methylpropene.

You should be able to see why this is called an elimination reaction.

HBr is removed (eliminated) from the haloalkane molecule  (It is product and should be in the equation).

The experiment can be carried out in the laboratory by use of a simple test tube apparatus collecting the gaseous alkene over water.

You should be able to draw the apparatus used as it is essentially the same kit as that used below in cracking liquid paraffin.

You should of course be able accurately to add labels to your diagram

You should be able to give the equation for the reaction of 2-bromobutane and ethanolic KOH in which two different alkenes are formed.

You should be able to name and draw the displayed formulae of these two alkene products.