# pls help me im so stuck!!

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#1
I'm really confused, ebery question I do on hess's law/enthalpy change I get it wrong. the markscheme for this question says: -642-286-(-602/2 x -92) which gives -142jkmol-1
BUT i got -1345kjmol-1, im so confused i used the enthalpy of formation equation= products-reactants because i really dont want to use those hess law diagrams, i did -286+(-642)-(-92 x 2)+(602) = -1345

but using the enthalpy of formation question how would ik i have to divide -602 by 2 and multiply -92 by 2 like im supposed to from the equation? pls help me

Attachment 736052
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2 years ago
#2
do you have a picture of the question ?
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2 years ago
#3
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#4
thanks for letting me know, hopefully its up now?
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#5
do you have a picture of the question ?
yeah sorry can u seee it now
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2 years ago
#6
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
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#7
(Original post by theowinn)
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.
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2 years ago
#8
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
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2 years ago
#9
thank u so much! so what i did was right but i have to work it out bit by bit.
Yeah man your good just make sure your signs are correct a two minus' becomes a plus remember!
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#10
(Original post by igotnomoney)
Stay calm and don't pit everything on calculator . . Do product separate and reactant separate .. good luck
thank u so much lol i just put everything on the calculator every time i tried these questions
1
#11
(Original post by theowinn)
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
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2 years ago
#12
thanks, do u mind helping me on another question, cba to set up another forum
Whats the question?
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#13
(Original post by dip0)
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
0
2 years ago
#14
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.
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#15
(Original post by dip0)
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.
wow thank u so much, so is there no way i can use the formula for these hess law types of questions, my teacher taught that u could use either of them, so i have to use the diagram in this case? also how do u know which route or way to go, which enthalpy change are we trying to find out? im guessing its the one that says enthalpychange in the diagram?
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