 How Far, How Fast?

Note: From 2008, there will be greater emphasis on "green" Chemistry and environmental issues.

This is only a half module, meaning the exam lasts 45 minutes and is marked out of 45, rather than 60. That means there's less to learn, but it's essential to get revising early!

Enthalpy Changes

You'll remember good old exothermic and endothermic reactions from GCSE. Exothermic is letting out heat into the surroundings in a reaction and endothermic is taking it in. There's a special name for this, and as you've probably guessed from the heading, it is enthalpy change.

The symbol for enthalpy is a H and the enthalpy change is given the two symbols ΔH (good old Greek letter Δ -- capital Delta).

If a reaction is exothermic, the ΔH value is negative. E.g. -164 kJ

If a reaction is endothermic, the ΔH value is positive. E.g. +164 kJ

Enthalpy change is given the units joules (J) or kilo joules (kJ). Remember, there are 1000 J in 1 kJ.

What makes reactions exothermic or endothermic, though? Well, a reaction is basically just a combination of breaking bonds, and then making bonds. We need to put in energy to break a bond, and energy is released when bonds are formed. If more energy is released than was put in, it becomes exothermic, if less energy is released, it's endothermic. Pretty simple. The MINIMUM initial energy you need to put into the reaction is called the activation energy.

Taking the above examples, we can draw diagrams of enthalpy change. Examiners like to ask you to sketch them, so make sure you know them!

 Exothermic Reaction Endothermic Reaction In order to compare enthalpy change, there has to be a standard for all experiments to follow. The symbol is used to denote standard conditions. These are:

Pressure: 1 atmos
Temperature: 25oC
Solution concentrations: 1 mol dm-3

So, the standard enthalpy change of a reaction -- written as ΔH -- just means the heat exchange with the surroundings of a reaction, under standard conditions.

The standard enthalpy change of combustion -- written as ΔHc -- means the complete combustion of ONE MOLE of a substance in O2 under standard conditions.
For example, for CH4: CH4 + 2O2 -----> CO2 + 2H2O

The standard enthalpy change of formation -- ΔHf -- means the formation of ONE MOLE of substance from its constituent elements under standard conditions.
For example, for CH4: C + 2H2 -----> CH4

And finally, the standard enthalpy change of Bond Dissociation -- ΔHd -- means then energy required to break one mole of a bond under standard conditions.
For example, for HCl: HCl -----> H + Cl (note, we have no charges here).

Back to the Top

Bond Enthalpy

Bond enthalpy is the energy required to break a covalent bond.

The bond enthalpy is written with the symbol E. For example, E(C-C) means the bond enthalpy of a C-C bond. The bond enthalpy of any bond is the same value as the ΔHd.

Some common bond enthalpies are:

 Bond E(X-Y) kJmol-1 H-H +/- 436 C-C +/- 347 C-H +/- 612 O=O +/- 413 O-H +/- 464 C-O +/- 358 C=O +/- 805

You don't need to remember those though, because all questions will give you the required info!

As you see from the table, bond enthalpy is measured in kJmol-1. It can also be either a positive (+ve) or negative (-ve) value. The sign is VERY important.

If you're breaking a bond, the bond enthalpy is positive (breaking bonds is endothermic and needs energy).

If you're forming a bond, the bond enthalpy is negative (forming bonds is exothermic).

The bond enthalpy is only for gaseous bonds, which makes measuring it, tricky. We use an AVERAGE bond enthalpy, which means calculations might vary from the real thing...

Using bond enthalpies, we can calculate the overall enthalpy change of a reaction.

From the above table, work out the enthalpy of combustion for CH4.

Before we do anything, we need to write down the equation. Combustion, remember, means burn completely in oxygen, so therefore, we get the equation:

CH4 + 2O2 -----> CO2 + 2H2O

Now, draw them out in their structural formula: Consulting the table above, we write down what we have. On the LHS of the arrow, all values are positive because the bonds are being broken, while on the RHS, they're negative because bonds are being formed.

 4 x + 612 = + 2448 2 x + 413 = + 826 2 x - 805 = - 1610 4 x - 464 = - 1856

Finally, add up all these values (remembering the negatives) and you'll get the answer:

- 192 kJmol-1

Easy! Remember your +/- !

Enthalpy Change of Solutions

Bond enthalpies are fine when we have gaseous molecules, but what happens with solutions? As you might remember, there were different mol equations and it's no different here. The equation for enthalpy change of solutions is:

ΔH = mcΔT

Don't look so confused!

ΔH is just the enthalpy change that we're calculating. It can be either +ve or -ve, so don't forget your sign! The answer to the equation comes out in joules - you merely divide by 100 to get kJ.

m is the mass, but we assume that 1 cm3 of solution is the same as 1 g. The mass is the final volume. (i.e. if you added two solutions together, you add those volumes).

c is a constant (gah, it's sounding like maths), which represents the heat capacity. This is usually around 4.2Jg-1K-1, but the exam question will always tell you what figure to use as the constant.

ΔH is the change in temperature. If the temperature falls, we attach a +ve (to show it's endothermic) and if the temp rises, we attach a -ve (to show it's exothermic).

Okay, a worked example to show how it's done:

50cm3 of concentrated sulphuric acid (10mol dm-3) was added to 950cm3 of water at a temperature of 19oC. The mixture was stirred gently until a maximum temperature of 28oC was reached. What was the enthalpy change for the reaction? Assume standard heat capacity is 4.2Jg-1K-1.

First off, we remind ourselves of the equation: ΔH = mcΔT

Next, it's just a case of sticking the numbers in:

m = 50 + 950 = 1000

c = 4.2

ΔT ΔT= 28oC - 19oC = 9oC. It's exothermic, therefore: -9.

Times all these numbers together and we get: -37800 J

Finally, divide by 1000 to get: -37.8 kJ

The question might ask you to calculate the enthalpy change per mole. This just means you divide the energy given out by the number of moles you have. In the example above:

50cm3 of concentrated sulphuric acid (10mol dm-3)

Remember the old formula: 50 x 10 1000 = 0.5 mol.

Then, - 37.8 kJ 0.5 = -75.6 kJmol-1

Easy!

Hess's Law

Hess was a scientist. He discovered:

"The enthalpy change of a reaction is independent to the route taken." Going from A + B to C + D is the same as going from A + B to E + F and then to C + D. From this, we can say: We can use Hess's Law to calculate enthalpy changes that are difficult to measure in the lab. We use data from two sets of data:

ΔHf -- the standard enthalpy change of formation -- means the formation of ONE MOLE of substance from its constituent elements under standard conditions.

ΔHc -- the standard enthalpy change of combustion -- means the complete combustion of ONE MOLE of a substance in O2 under standard conditions.

Examples

Calculate ΔH for CH4 + 2O2 ---> CO2 + 2H2O from the following data:

 Compound ΔHf /kJmol-1 CH4 - 76 CO2 - 394 H2O - 286

So, we have ΔHf here, so we need the following diagram: Because O2 is an element, it has no formation value. Because there are 2 moles of H2O, we need to times the value by 2. -286 x 2 = - 572.

We want to go from CH4 and O2 down to the constituent elements and up to the products, so we need to change the arrows, and thus change the signs: Finally, we add up all the numbers, being careful of the plus and minus signs and get the answer: - 890 kJ mol-1

Calculate the ΔHc for 2C + 2H2 + O2 ---> CH3CO2H, using the following data:

 Compound ΔHc /kJmol-1 C -394 H2 -286 CH3CO2H -876

This time we have ΔHc values, so the arrows go the opposite way: We want to go from C, H2 and O2 and then up to CH3CO2H.

So we change the arrow, and the sign. Add up to get the answer: -484 kJ mol-1

That's all there is to it! Remember, fomation arrows go from the box, combustion arrows go down to the box!

Back to the Top

Pollution

Vehicles powered by an internal combustion engine are major polluters. Catalytic convertersare used to decrease the amount of pollution.

Catalytic converters cylindrical structures containing a ceramic honeycomb coated in a mixture of palladium, platinum and rhodium. The honeycomb structure gives the maximum surface area so that the catalyst can be more effective, and it is fixed on the exhaust, so that the hot gases will react as they pass through.

There are three main pollutants that you should know about:

 Hydrocarbons Carbon Monoxide Oxides of Nitrogen Formation Some of the fuel escapes unburned. When hydrocarbons burn with insufficient O2, CO is produced. O2 and N2 combine in heat of exhaust. Effect Carcinogenic and toxic. CO is highly toxic. It combines with haemoglobins in the blood to prevent oxygen reaching tissue. Nitrogen oxides cause smog, low level ozone, are irritant and contribute to acid rain. Catalytic Converter Oxidises the hydrocarbon into CO2 and H2O. Oxidises CO to CO2 and H2O: Step 1: Adsorption of CO onto active site of catalyst. Step 2: Bonds are weakened so chemical reaction occurs. Step 3: Desorption of CO2 and H2O from the catalyst. Reduces nitrogen oxides N2: Step 1: Adsorption of NO onto active site of catalyst. Step 2: Bonds are weakened so chemical reaction occurs. Step 3: Desorption of N2 from the catalyst.

Catalytic converters aren't the complete answer, however, because CO2 is one of the main contributors to global warming. We'd be so much better off if we walked or cycled...

Back to the Top

Collision Theory

In order to react, particles need to collide with the minimum activation energy and the correct orientation.

If we increase the pressure of a gas, or the concentration of a liquid, we increase the rate of reaction. This is because molecules have less space to move around in and therefore are more likely to collide. A greater number of collisions means a greater chance of colliding with the EA and correct orientation as so reacting.

If we increase the temperature, the rate of reaction increases again. This is because of two factors. Providing more heat gives the particles more energy, and so they move around faster. Thus they are more likely to collide and more collisions means a greater chance of a reaction. Also, with more energy, particles are more likely to have the correct EA and so when they collide, there's an even greater chance of reaction.

We can see the affect of temperature in the Boltsman distribution.

In a sample of gas, the Boltsman distribution shows the different energies of particles: • The area represents the total number of particles. For a higher temperature, there are a lot more particles with energy above the EA.
• No particles have no energy, so the graph starts at the origin.
• The graphs never touch the x-axis again, though, because there is no limit to the amount of energy a particle can have.

A catalyst also increases the rate of reaction. It enters a reaction, provides an alternate route for the reaction to take place and so lowers the activation energy, without being consumed.

Labelling a catalyst on the Boltsman distribution would look like this: There are two types of catalyst, homogeneous and heterogeneous.

A homogeneous catalyst is in the same phase as the reactants, such as sulphuric acid in the esterification of an alcohol: All these are liquid.

Another example is the chlorine free radical breakdown of the atmosphere:

```                         Cl + O3 -----> ClO + O2
ClO + O -----> Cl + O2```

Overall equation: O3 + O -----> 2O2

All these are gaseous.

A heterogeneous catalyst is in a different phase to the reactants.

Catalytic converters (palladium, rhodium and platinum) are a good example.

Iron in the Haber Process is also a heterogenous catalyst:

N2(g) + 3H2(g) 2NH3(g)

Back to the Top

Chemical Equilibrium

A chemical equilibrium occurs when a forward and backward reaction are moving at the same speed. It takes place in a closed system and because the rates of the forward and reverse reactions are equal, the concentrations are fixed.

Controlling a chemical equilibrium is really useful in industry.

According to Le Chatelier's Principal, whenever a dynamic equilibrium is disturbed, it changes so that the disturbance is opposed and equilibrium is restored.

Temperature/pressure/concentration can all be altered to produce more or less product. The production of ammonia in the Haber Process is a good example of using equilibriums.

The Haber process is very important because ammonia used to make fertilisers, such as ammonium sulphate:

NH3 + H2SO4 -----> (NH4)2SO4 + H2O

Ammonia is also used to produce nitric acid, which is used in explosives.

The equation that occurs in the Haber Process is:

 N2(g) + 3H2(g) 2NH3(g) ΔH = - 72 kJ mol-1

If we look at pressure, we can see that there are 4 moles on the LHS, but only 2 moles on the RHS. This means that if we increase the pressure, due to Chetalier's Principal, the equilibrium will try to oppose this. There will be more product produced, because on the RHS, there are fewer moles. This is what we want. However, increasing pressure is very expensive...

The same happens with concentration.

If we consider temperature, however, the ΔH = - 72 kJ mol-1. This means: N2(g) + 3H2(g) 2NH3(g) It's endothermic towards the reactants side, and so if the temperature is increased, the equilibrium will try to oppose it by moving to the LHS and produce less product. This isn't what we want.

However, a decreased temperature will slow the rate of reaction, and thus also lessen the yield...

In real life, a compromise is reached. The conditions for the Haber Process are around 400oC and 250 atmos.

Back to the Top

Acids and Bases

From GCSE, you'll have been told that acids had a pH below 7, alkalis a pH over 7. As usual, they were telling you lies and we now have a new definition for acids and bases:

Acids are proton donors.

Bases are proton acceptors.

What does that mean? Well, when in aqueous form, acids dissociate into their ions, and one of these is a H+.
Example: HCl --> H+ + Cl-
If a hydrogen loses its electron, it basically becomes a proton. When they react, the base takes the H+ ion, "accepting" it, while the acid "donates" it.

If we look at the following equation:

HCl + HBr ---> H2Cl- + Br-

We're used to HCl being an acid, but here, it is accepting a H+ and so is the base. This is because HCl is a weaker acid than HBr.

Strong and Weak Acids

We can differentiate between strong and weak acids depending on their ionisation or dissociation.

Strong acids fully dissociate, which means in aqueous form they split completely into their ions. HCl is an example:

HCl ---> H+(aq) + Cl-(aq)

Not all acids fully dissociate fully, however. Weaker acids only partially dissociate into ions and there are still complete molecules. Organic acids, like methanoic acid are an example:

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

There is a lower concentration of H+ ions and this means the reaction is slower.

Main Reactions - just gotta learn some, I'm afraid...

1. Acid + Metal -----> Salt + Hydrogen
Equation: Ca + H2SO4 -----> CaSO4 + H2
Ionic Equation: Ca + 2H+ ------> Ca+2 + H2
2. Acid + Carbonate -----> Salt + Water + Carbon Dioxide
Equation: Na2CO3 + H2SO4 -----> Na2SO4 + H2O + CO2
Ionic Equation: CO3-2 + H2+ -----> H2O + CO2
3. Acid + Base -----> Salt (a neutralisation reaction)
NH3 + HCl -----> NH4Cl
NH4+ + Cl- -----> NH4Cl
4. Acid + Alkali -----> Salt + Water (another neutralisation reaction)
HCl + NaOH -----> NaCl + H2O
H+ + OH- -----> H2O
Back to the Top

That's just about everything for How Far, How Fast!