pls help me im so stuck!!

do you have a picture of the question ?

thanks for letting me know, hopefully its up now?

The Standard Enthalpy change is:

Sum of enthalpy of formation i.e /\H f (Products) - Sum /\H (reactants).
(-286-642) - (-602- (2x92)) = -928+786= -142KJmol^-1

thank u so much! so what i did was right but i have to work it out bit by bit.

thank u so much! so what i did was right but i have to work it out bit by bit.

Stay calm and don't pit everything on calculator . . Do product separate and reactant separate .. good luck

Yeah man your good just make sure your signs are correct a two minus' becomes a plus remember!

thanks, do u mind helping me on another question, cba to set up another forum

Whats the question?

how would i do this question without using the hess law diagrams, would i use the enthalpy of formation=products-reactants
the markscheme shows
ΔH + 963 = –75 – 432 OR ΔH + 963 = – 507 (M1)
ΔH = –75 – 432 – 963 (M1 and M2)
ΔH = –1470 (kJ mol–1) Award 1 mark for + 1470

but i dont understand this markscheme

Whenever you are given enthalpy data, the only way about to answering the question is to construct a Hess's law cycle, there is no other way around.
You are familiar with the definition of Hess's law right?

We use -963 instead of +963 because we are interested in the opposite reaction going, which means this is an exothermic process as opposed to the reaction (+963) given in question which is endothermic.

As for what the MS is trying to say, think of it like this:
if I wanted to go from Birmingham to Manchester, but couldn't go directly. Then I could go from Birmingham to Nottingham, then from Nottingham to Manchester. If the displacement from Birmingham to Nottingham is X, then the displacement from Nottingham to Birmingham is -X (going in opposite direction).
Hope this analogy is of some help.