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Hess's law help me!

4 (c) Dinitrogen oxide is formed when ammonia is oxidised according to the following
equation.
2NH3(g) + 2O2(g) N2O(g) + 3H2O(l)
4 (c) (i) Use the standard enthalpies of formation in the table below to calculate a value for the
standard enthalpy change of this reaction.

............ΔHf / kJ mol–1
NH3(g)= -46
O2(g)= 0
N2O(g) = +82
H2O(l)= -286



This is what i did:
46 x 2= 92

-286 x 3= -858
-858 + 82= -776

ΔH = ΣΔH products ΣΔH reactants

-776-92= -864

The MS says that the answer is actually -686.
what did i do wrong?
(edited 9 years ago)
Reply 1
-(2x-46)+(+82)+(3x-286) = -684 :frown:

You missed a minus sign (46 x 2= 92)
Original post by Pigster
-(2x-46)+(+82)+(3x-286) = -684 :frown:

You missed a minus sign (46 x 2= 92)


but i thought i had to change the sign? ( thats what it says in the nelson thornes aqa book)
Reply 3
You changed it twice, once when you did 46 x2 and again when you did products minus reactants.
I got -684 too
Is the mark scheme wrong?
Original post by Pigster
You changed it twice, once when you did 46 x2 and again when you did products minus reactants.

ok so i should have just put all the numbers in the equation without changing the signs :
ΔH = ΣΔH products ΣΔH reactants

so the equation basically changes the signs for me ?
Original post by EliteHoneyBadger
ok so i should have just put all the numbers in the equation without changing the signs :
ΔH = ΣΔH products ΣΔH reactants

so the equation basically changes the signs for me ?


Yeah
Original post by Legal drugdealer
Yeah

oh thanks a lot for all your help :smile:
Original post by EliteHoneyBadger
oh thanks a lot for all your help :smile:


no problem :smile:
Original post by EliteHoneyBadger
but i thought i had to change the sign? ( thats what it says in the nelson thornes aqa book)

82+3(-286)-2(-46) = -684

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