Chemistry Qs

I need to know more type of these Qs and practice them as I am not familiar with combining enthalpy and how they did it and where is the 2?
What is this call when there is two enthalpy reactions?
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-H/QP/Enthalpy%20Changes%20QP.pdf
Q3 I think.

For Q3, you need to be aware of something called a Hess cycle.
The Hess cycle in this example is a bit of a weird one. The reactions you have to work with are all given and the data to work out the enthalpy change for
Na2O(s) + H2O(l) —> 2NaOH(aq)
is given.
Your best starting point is to use these data to calculate ΔH for this reaction.
Essentially you need to fit the three reactions you are given into a Hess cycle. This sometimes means adding extra species to each side of the balanced equations that ultimately don’t actually take part in the reaction.
I’d make the top line
Na2O(l) + 2HCl(aq) + H2O(l) —> 2NaCl(aq) + 2H2O(l)
The extra water molecule (that doesn’t participate in the overall reaction) was added because one reaction that you are given is Na2O(s) + H2O(l) —> 2NaOH(aq) and so you need to throw water in somehow to be able to even include this equation in the Hess cycle.
I’d make the bottom line
2NaOH(aq) + 2HCl(aq)
The arrow from the reactants to the products on the top line should be labelled ΔrH° or similar. This is the value you need to find.
The arrow from the bottom line to the products should be labelled 2 x -57.6 kJ/mol, since the bottom line shows 2 x the number of reactants given in the equation for the reaction
NaOH(aq) + HCl(aq) —> NaCl(aq) + H2O(l) ΔrH° = -57.6 kJ/mol
The arrow from the reactants to the bottom line will need to be labelled with the enthalpy change worked out from the data.
If you recall vectors from GCSE maths, you can treat a Hess Cycle similarly. You could either directly react Na2O(s) and 2HCl(aq) to get to your products, or alternatively you could react Na2O(s) and H2O(l) to make 2NaOH(aq) first and then react the NaOH(aq) with HCl(aq) to get the same overall reaction. If you think about it this way, the overall enthalpy change is therefore the sum of the given enthalpy change x 2 and the calculated enthalpy change.

If you look at the very last part. I don’t think they teach you enthalpy change of vaporisation in AS but I used the same logic. But what I don’t understand is why is enthalpy change of vaporisation of water is being multiplied by - how do you know it’s exo ? Tell me some must and key ideas about enthalpy change of vaporisation I been through almost all of it?

Okay, so 7(b), then.
Well, the standard enthalpy change of combustion of an organic compound will form CO2(g) and H2O(l).
Essentially your Hess cycle should have the overall reaction on the top line and the bottom line should be 3CO2(g) + 4H2O(l).
The arrow from the bottom line to the products will be 4 x ΔvapH [H2O(l)] (because there are 4 moles of water molecules and the value they give you is for 1 mole of water molecules) and the arrow from the reactants to the bottom line will be the standard enthalpy change of combustion of propane.
You can work out ΔH for the overall reaction using the bond enthalpies (recall that bond enthalpies are measured for compounds in the gas phase) and so the only thing you are missing is the standard enthalpy change of combustion.
The enthalpy change of combustion for any reaction will always be exothermic. This is because without fail more bonds are formed than broken.

For enthalpy change of formation it’s endo ?
For enthalpy change of reaction it’s either one?
For enthalpy change of vaporisation it’s exo?

Enthalpy changes of formation can be either endothermic or exothermic - usually they are exothermic, but unstable compounds often have endothermic formation enthalpies.
Enthalpy changes of reaction can be either. It depends on the formation enthalpies of all the relevant species.
Enthalpy changes of vaporisation are always endothermic. You can think of them as being a measure of how much energy (per mole) you need to input in order to overcome the intermolecular forces holding the molecules in a liquid together.
I’m wondering if the signs are confusing you. Positive enthalpy changes are endothermic processes and negative enthalpy changes are exothermic processes.

For this one https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Enthalpy%20Changes%201%20QP.pdf
I don’t understand why you need the moles to work out enthalpy change for 5bi.
Also I completely don’t understand the logic and visualisation for 5b(iii). Tell me some scenarios as I can’t visualise this.

One important feature of a reversible reaction is that the enthalpy changes of the forward and backward reactions have the same magnitude, but opposite signs (e.g one is endothermic and one is exothermic).
If we work out the reaction profile for the reverse route of the reaction (which is exothermic), it becomes far easier to spot.

But why equivalent the equation for 5bi. Is there a need to anyway if you could use just one equation that’s it?
In chemistry is there a special name for chemical equations that are the same?

The equivalent one is better to work with because you show the formation of just one mole of the product of interest and therefore scale down the enthalpy change to reflect it. Because you are making 10 moles of N2O(g), it’s easier to spot how to scale it correctly if you are working with an equation that shows the formation of just one mole of the product rather than two.
As for your second question, I am unsure exactly what bit you are referring to.

For Q(Iv) they give us data for combustion. I use the general formula for combustion and obtain it that alway but how would I obtain for a formation value?

The equivalent one is better to work with because you show the formation of just one mole of the product of interest and therefore scale down the enthalpy change to reflect it. Because you are making 10 moles of N2O(g), it’s easier to spot how to scale it correctly if you are working with an equation that shows the formation of just one mole of the product rather than two.
As for your second question, I am unsure exactly what bit you are referring to.

The equivalent one is better to work with because you show the formation of just one mole of the product of interest and therefore scale down the enthalpy change to reflect it. Because you are making 10 moles of N2O(g), it’s easier to spot how to scale it correctly if you are working with an equation that shows the formation of just one mole of the product rather than two.
As for your second question, I am unsure exactly what bit you are referring to.

Also I don’t get 5(iv) how are am I suppose to know about the bond strength. All I know is that sigma bonds are stronger than pie bonds can’t form simultaneous equations. Assuming you divide by 2?

For Q(Iv) they give us data for combustion. I use the general formula for combustion and obtain it that alway but how would I obtain for a formation value?

For Q1C how do you write the equation of formation for the equation that shows combustion and how to show the formation in Hess law cycle as a diagram . I got it right by using the formula equation just curious how to do it that way?
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Enthalpy%20Changes%204%20QP.pdf

The enthalpy of formation of glucose should be
6C(s) + 6H2(g) + 3O2(g) —> C6H12O6(s)
All you do is you make the top line the enthalpy of combustion of glucose and the bottom line the reactants in the equation above + 6 more molecules of O2(g)