# A-level Chemsistry- Hess' Law

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#1
Calculate the standard enthalpy change for the following reaction using the enthalpy changes given
SO ₂(g)+2H ₂S(g)------------>3S(s)+2H ₂O(l)
ΔH°c: S(s): -297KJmol-1
ΔH°f: H ₂O (l):-286 KJ mol-1
H ₂S (g):-20 KJ mol-1
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1 year ago
#2
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1 year ago
#3
I could give you the exact answer, but it's against community guidelines. It's not actually Hess' law. You have to use the sum of all the moles of products-the sum of all the moles of reactants
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#4
(Original post by JIMBO789)
I could give you the exact answer, but it's against community guidelines. It's not actually Hess' law. You have to use the sum of all the moles of products-the sum of all the moles of reactants
I know the exact answer but I just don't know how to get to it
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1 year ago
#5
(Original post by he3456)
I know the exact answer but I just don't know how to get to it
You weren't given the SO2's standard enthalpy?
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1 year ago
#6
Draw a Hess cycle

SO ₂(g)+2H ₂S(g)------------>3S(s)+2H ₂O(l)

write S + 2H2 + O2
draw an arrow from S + 2H2 + O2 to the right hand side labelled as 2ΔH°f H ₂O
draw an arrow from S + 2H2 + O2 to the right hand side labelled as ΔH°c S(s) + 2ΔH°f(H2S)

hence ΔH°c[S(s)] + 2ΔH°f(H2S) + ΔHr = 2ΔH°f(H₂O)
solve for ΔHr = -235 kJ/mol
1
1 year ago
#7
Enthalpy of formation of SO2 is the same as enthalpy of combustion of S. The relevant reaction is
S + O2 -> SO2.
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#8
(Original post by BobbJo)
Enthalpy of formation of SO2 is the same as enthalpy of combustion of S. The relevant reaction is
S + O2 -> SO2.
But using this gives me an answer of 576
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1 year ago
#9
(Original post by he3456)
But using this gives me an answer of 576
oh ok thx
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