you would be right if it said that but it said formation there was delta Hf not delta Hr
I'd rather not argue the toss but it said neither of those things, The question said, word for word: 'Find the standard enthalpy change for this reaction'
"Standard enthalpy change of reaction, ΔH°r - The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state."
^ The question asked for the standard enthapy change of the reaction given and you were given the correct ratio for the reaction stated, which was 2 SO2 + O2 -> 2SO3 and the standard enthalpy of formation values for these substances so you were expected to use D.H = D.H(Products) - D.H(Reactants) to work out the enthalpy change for the reaction given. The D.H calculated would give you the standard enthalpy change of such a reaction under standard conditions, which is what the question asked for.
Then you worked out the entropy change for the reaction given and put the figures into the gibbs free energy equation to work out the free energy change for the reaction given.
tl;dr: there was no reason to divide by two, if you didn't you are correct.
Definition when I googled standard enthalpy change:
The standard enthalpy of reaction (denoted ΔHr⊖) is the enthalpy change that occurs in a system when one mole of matter is transformed by a chemical reaction under standard conditions.
I don't know anymore! Haha
Nice back-peddle, I suggest you read up on your definitions because that is wrong:
I got something like 2328 KJ Mol-1 at first, then I halved the atomisation of oxygen to get 2404, but I think the original answer of 2328 was correct, because atomisation is for one mole, and you got 1 mole of O
The question asked 'for this reaction' - this is the standard reaction that occurs when OXYGEN is in its standard state (O2) is the one that was given.
Halving the values would be meaningless and would not be for the standard reaction when reagents and products are in their standard states.
Would saying increase in pressure increase emf value, equilibrium shifts to RHS be ok for 2 marks do you think?
Thanks.
SA has no effect as it referred to BOTH platinum electrodes.
Yes, emf did increase at greater pressure and your reasoning is correct, although I think I worded it slightly differently with reference to the p.d. produced in a given time with an increased reaction rate.
The reaction in the contact process is not delta Hf, SO2 and O2 are not 'elements' It is delta Hr, there is no need to divide by 2, in fact technically it is wrong to divide by 2
As a simple analogy: For this reaction A + B -> C I'm sure you'd be fine with this being standard About this one: A + B -> C + 2D Well C is 1 mol, about D?
That's where the argument of 1 mol breaks down. In fact, standard enthalpy of reaction specifically does not specify 1 mol for that very reason - because it's incorrect
I get your point but in the question it stated delta H f because I was working out it without halving till I saw that it had the symbol there was no definition on the paper and you had to apply your knowledge of that why would they state delta Hf if they didn't want one mol
Explain to me how this is any different to the question. You're provided with enthalpy of formation values and asked to work out the standard enthalpy change of the following reaction exactly as we did.
Just telling you now, there's absolutely no diving in the mark scheme.
SA has no effect as it referred to BOTH platinum electrodes.
Yes, emf did increase at greater pressure and your reasoning is correct, although I think I worded it slightly differently with reference to the p.d. produced in a given time with an increased reaction rate.
Excellent - my thoughts exactly!
Ok I might deduct a mark for my explanation then, thanks