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# Thermochemical cycles/Hess's Law help watch

1. Hi,

I've been doing some AS revision and I'm really struggling with drawing the cycles. I'd be really grateful if someone could guide me in the right direction as I'm not sure what I'm doing wrong. Thanks

Q6. I've tried drawing the cycle but I think I've made a few mistakes as I can't get to the right answer.

Q5. I'm really not sure which bit goes where.
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2. 5) The first thing you need to do is look at the reactions you're given - reactions 1 and 3 are both going in the same direction as they consume similar reactants and produce similar products, but reaction 2 is the wrong way around, so you will need to use the reverse, which will have the opposite enthalpy change:
TiO2(s) -> Ti(s) + O2(g) ΔH = 912kJmol-1

After that, I would look at how reactions 2 and 3 can be used to explain the steps of reaction 1.
Starting with C(s) + TiO2(s) + 2Cl2(g), the first thing that could happen is the TiO2 decomposes into Ti(s) and O2(g), so we end up with C(s) + Ti(s) + O2(g) + 2Cl2(g).

The question is asking about a standard enthalpy of formation, so we need to look for elements in their standard states. If you look at the list of products from that last reaction, you can see that we now have all of the elements in this form! To get to the products of reaction 1, we would first need to convert C(s) + O2(g) -> CO2(g) (we are given this ΔH), and then Ti(s) + 2Cl2(g) -> TiCl4(l) (this is the ΔHf of TiCl4).You could write this in a cycle like this:

.
6) Look at your balanced equation, you have 5H on the left but only 2H on the right. You have exactly the right idea and your intermediate step with 5/2H2 is correct, I think you just forgot to carry this on to the final reaction.
3. (Original post by lizardlizard)
5) The first thing you need to do is look at the reactions you're given - reactions 1 and 3 are both going in the same direction as they consume similar reactants and produce similar products, but reaction 2 is the wrong way around, so you will need to use the reverse, which will have the opposite enthalpy change:
TiO2(s) -> Ti(s) + O2(g) ΔH = 912kJmol-1

After that, I would look at how reactions 2 and 3 can be used to explain the steps of reaction 1.
Starting with C(s) + TiO2(s) + 2Cl2(g), the first thing that could happen is the TiO2 decomposes into Ti(s) and O2(g), so we end up with C(s) + Ti(s) + O2(g) + 2Cl2(g).

The question is asking about a standard enthalpy of formation, so we need to look for elements in their standard states. If you look at the list of products from that last reaction, you can see that we now have all of the elements in this form! To get to the products of reaction 1, we would first need to convert C(s) + O2(g) -> CO2(g) (we are given this ΔH), and then Ti(s) + 2Cl2(g) -> TiCl4(l) (this is the ΔHf of TiCl4).You could write this in a cycle like this:

.
6) Look at your balanced equation, you have 5H on the left but only 2H on the right. You have exactly the right idea and your intermediate step with 5/2H2 is correct, I think you just forgot to carry this on to the final reaction.
Thank you so much for your help I've managed to get to the right answers now!
4. Hey! Just wondered if you could help, this is a question relating to lattice enthalpy problems

Why are ionic compounds with some covalent character insoluble in water?

Thank You

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