Chemistry Reviews (AS level)

Chapter 9: Enthalpy

Standard concentration- 1moldm⁻³

Standard state- whatever the state of the element/molecule is at standard conditions (so at room temp)

If enthalpy changes aren't found with standard conditions, it could result in inaccurate results. There are also other reasons for inaccuracies, like:

-Heat loss to the surroundings

-Incomplete combustion

-Evaporation

The bottom 2 are related to an experiment. It is a required practical, so I think you are supposed to know it, but usually in a test they explain the practical. Just in case, I'll give a brief desc. Basically a spirit burner filled with a specific fuel is lit, and heats up water for a set temperature change. Then, an equation (stated below) is used to find the heat energy transfer. Evaporation of the water would happen if you let the water heat to over boiling temp. Heat loss is reduced with draught shields, but it can still happen. 

That all leads us onto an equation. The unfortunate part is that we have to actually remember this one. Here it is: q=mc∆T. In that equation, m is mass, c is specific heat capacity (of the surroundings- could often be water), ∆T is change in temp, and q is the energy. I don't know why energy is q, I don't think there is anything wrong with E, but we have q just for this one. Maybe it is because it is heat energy, idk. But that equation is used to find the energy change in a reaction. If you struggle to remember that, here is something that kind of helps me: the equation reminds me of kumquat (it's a fruit). You could see the equation like the word, maybe like qmct (q=mc∆T. You could also rearrange it to make it cmqT (cm=q/∆T). Hope that helped.

Now for what this has all been leading up to: Bond enthalpies. This is to do with the breaking and making of bonds, and with these you can work out the overall enthalpy change of an equation. To work this out you are given average bond enthalpies of each bond, e.g. C-C  has an average bond enthalpy of 413kJmol⁻¹. Do keep in mind this is an average, so the actual bond enthalpy might be slightly bigger and smaller, depending on the overall molecule. 

Also, we use the thermodynamic knowledge from earlier here. Breaking bonds is endothermic, so has a +∆H. Making bonds is exothermic, so has a -∆H. With this, you can work out enthalpy changes. A good starting point is to look at the displayed structure, and count all the different bonds. Then, add them up for each side. Remember, breaking bonds is endo, so the reactants' enthalpy will be +, and making is exo, so the products' enthalpy is -. Then, do maths.

Time to move onto the next part, which does relate to the last section, as it is to do with enthalpies of a reaction (although I guess that is basically the whole topic). Don't be hessitant, it is time to ride the Hess cycle. Hess' law says that the enthalpy change of a reaction is the same regardless of the route taken. This means that even if a reaction happens indirectly, so other reactions happen before you get the products, the enthalpy change is the same. This is easier pictured, as you can see the triangle, but I'll try explaining with words.

First though, we need to go through the different types of enthalpy change. I skipped this from earlier to explain it all together. The first which we've already covered, is enthalpy of reaction, AKA ∆ᵣH°. This is just the overall enthalpy of an equation, which I have (possibly incorrectly) said as ∆H earlier. Same thing, probably. Also remember this is enthalpy of reaction per mole. This means that if the question is to find the ∆ᵣH° of a reaction which forms a mole of product, then you need to balance the equation to have a mole of the product, even if all the reactants have fractions have fractions of moles and it would seem to make more sense if you had whole numbers. However, if you multiply that, you will also be multiplying the answer, so it will be wrong.