# Enthalpy change question

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(iii) Calculate the standard enthalpy change of formation, ∆fH ө, of hydrogen sulfide using the enthalpy change for Reaction 1, and the standard enthalpy changes of combustion below.

Substance ∆cH ө / kJ mol−1
S(s) −296.8
H2(g) −285.8

2H2S(g) + 3O2(g) --> 2SO2(g) + 2H2O(l) ∆rH = –1125 kJ mol−1 Reaction 1

S(s) + H2(g) ------------- > H2S(g)
\ /
\ /
> combustion products <

∆fH of H2S = -582.6 + 562.5 = - 20.1 KJmol-1

At first I got -40.2 KJmol-1, unfortunately i don't have the mark scheme.
(edited 3 years ago)
Original post by Orchard2
hey, hope you're doing well!
I am not sure about the answer, can you check out please?

thank you!

(iii) Calculate the standard enthalpy change of formation, ∆fH ө, of hydrogen sulfide using the enthalpy change for Reaction 1, and the standard enthalpy changes of combustion below.

Substance ∆cH ө / kJ mol−1
S(s) −296.8
H2(g) −285.8

2H2S(g) + 3O2(g) --> 2SO2(g) + 2H2O(l) ∆rH = –1125 kJ mol−1 Reaction 1

S(s) + H2(g) ------------- > H2S(g)
\ /
\ /
> combustion products <

∆fH of H2S = -582.6 + 562.5 = - 20.1 KJmol-1

At first I got -40.2 KJmol-1, unfortunately i don't have the mark scheme.

If given enthalpies of combustion, formula is: (sum of reactants) - (sum of products).
Original post by Orchard2
hey, hope you're doing well!
I am not sure about the answer, can you check out please?

thank you!

(iii) Calculate the standard enthalpy change of formation, ∆fH ө, of hydrogen sulfide using the enthalpy change for Reaction 1, and the standard enthalpy changes of combustion below.

Substance ∆cH ө / kJ mol−1
S(s) −296.8
H2(g) −285.8

2H2S(g) + 3O2(g) --> 2SO2(g) + 2H2O(l) ∆rH = –1125 kJ mol−1 Reaction 1

S(s) + H2(g) ------------- > H2S(g)
\ /
\ /
> combustion products <

∆fH of H2S = -582.6 + 562.5 = - 20.1 KJmol-1

At first I got -40.2 KJmol-1, unfortunately i don't have the mark scheme.

-296.8 - 285.8 + 1125/2 = -20.1 kJ
Original post by Orchard2
hey, hope you're doing well!
I am not sure about the answer, can you check out please?

thank you!

(iii) Calculate the standard enthalpy change of formation, ∆fH ө, of hydrogen sulfide using the enthalpy change for Reaction 1, and the standard enthalpy changes of combustion below.

Substance ∆cH ө / kJ mol−1
S(s) −296.8
H2(g) −285.8

2H2S(g) + 3O2(g) --> 2SO2(g) + 2H2O(l) ∆rH = –1125 kJ mol−1 Reaction 1

S(s) + H2(g) ------------- > H2S(g)
\ /
\ /
> combustion products <

∆fH of H2S = -582.6 + 562.5 = - 20.1 KJmol-1

At first I got 40.2 KJmol-1, unfortunately idon't have the mark scheme.
please is that the answer I want to be sure
You can draw out a Hess Cycle if it helps.

2H2S(g) + 3O2(g) --> 2SO2(g) +2H2O(l)
\ / /
2S(s), 2H2(g), 3O2(g)

Filling in the relevant enthalpy changes and calculating will yield you -20.1kJ mol-1, which is the correct answer.
You may have got -40.2kJ mol-1 as your answer if you forgot that it makes 2 moles of H2S, so of course you simply need to divide by 2 for your final answer.
(edited 7 months ago)