Given : S(s) + O2(g) --> SO2 (g) Delta H= -296.1 kJ 2SO3(g) --> 2SO2(g) + O2(g) Delta H= 198.2 kJ How to find the Delta H for :2S(s)+3O2(g) --> 2SO3(g)?

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Answer 1 · 2018-03-15T17:02:55

We want $DeltaH^@$ for..

$2S(s) + 3O_2(g) rarr 2SO_3(g)$

We gots...

$(i)$ $S(s) + O_2(g) rarr SO_2(g)$ $;DeltaH^@=-296.1*kJ*mol^-1$

And...

$(ii)$ $2SO_3(g) rarr2SO_2(g) + O_2(g)$ $;DeltaH^@=+198.2*kJ*mol^-1$

And we take $2xx(i)-(ii)$ ...

$2S(s) + 2O_2(g) +cancel(2SO_2(g)) + O_2(g)rarr2SO_3(g)+ cancel(2SO_2(g))$

.....the which gives...

$2S(s) + 3O_2(g) rarr2SO_3(g)$

$DeltaH_"rxn"^@=(2xx-296.1-198.2)*kJ*mol^-1=-790.4*kJ*mol^-1$