Question

Given the data 2 S(s) + 3 O2(g) → 2 SO3(g) ΔH = −790 kJ S(s) + O2(g) → SO2(g) ΔH = −297 kJ SO3(g) + H2O(l) → H2SO4(l) ΔH = −132 kJ use Hess's law to calculate ΔH for the reaction 2 SO2(g) + O2(g) → 2 SO3(g).

Answer 1

2 S(s) + 3 O2(g) → 2 SO3(g) ΔH = −790 kJ

S(s) + O2(g) → SO2(g) ΔH = −297 kJ

SO3(g) + H2O(l) → H2SO4(l) ΔH = −132 kJ

Get

2 SO2(g) + O2(g) → 2 SO3(g)

NOTE: do not add equation 3, we can't cnacel H2SO4

only use 1 and 2

2 S(s) + 3 O2(g) → 2 SO3(g) ΔH = −790 kJ

S(s) + O2(g) → SO2(g) ΔH = −297 kJ

invert (2)

2 S(s) + 3 O2(g) → 2 SO3(g) ΔH = −790 kJ

SO2(g) →S(s) + O2(g) ΔH = +297 kJ

Mulitlpy (2) by 2

2 S(s) + 3 O2(g) → 2 SO3(g) ΔH = −790 kJ

2SO2(g) →2S(s) + 2O2(g) ΔH = 2*(297) = 594 kJ

add all equations

2 S(s) + 3 O2(g) + 2SO2(g) → 2 SO3(g) + 2S(s) + 2O2(g)   ΔH = −790 + 594 kJ = -196 kJ

cancel common terms

O2(g) + 2SO2(g) → 2 SO3(g)    ΔH = -196 kJ

its the same as

2 SO2(g) + O2(g) → 2 SO3(g)

then    ΔH = -196 kJ