how in the world do you do this problem?

Question

Use the thermochemical equations shown below to determine
the enthalpy (kJ) for the reaction:

H2SO3(l)=>H2S(g) + 3/2O2(g)

H2SO3(l)=>H2O(l) +SO2(g) DH=62KJ

SO2(g)=>S(s) + O2(g) DH=297KJ

H2S(g) +1/2O2(g)=>S(s) + H2O(l) DH=-155KJ

Answer 1

you need to add or subtract the rxns with the KJ given to create the rxn that you are trying to solve for.

H2SO3(l)=>H2O(l) +SO2(g) DH=62KJ

see that H2SO3 is common to the rxns, so you add this (that is, 62KJ)

so you have

H2SO3(l)=>H2O(l) +SO2(g)

as your rxn. Now we need to fix the product side...we want H2S(g) + 3/2O2(g)

use

SO2(g)=>S(s) + O2(g)

by adding this rxn to the first you get

H2SO3(l)+ SO2(g) =>H2O(l) +SO2(g) + S(s) + O2(g)

and (62KJ +297KJ)

you cross out anything common to both sides of the rxn so the equation becomes

H2SO3(l) =>H2O(l)+ S(s) + O2(g)

Now you need to use the last rxn to get the correct eqn. Sice we want the reactants of this rxn on the product side we need to reverse it (so we will subtract its KJ value from the eqn we already have => 62Kj+297kj-(-155Kj) )

H2SO3(l) + S(s) + H2O(l)=>H2O(l)+ S(s) + O2(g)+ H2S(g) +1/2O2(g)

cross out common..

H2SO3(l) => O2(g)+ H2S(g) +1/2O2(g)

add oxygens on product side

H2SO3(l) => H2S(g) +3/2O2(g)

62Kj+297kj-(-155Kj) =enthalpy of rxn

note:for future if you need 2,3,4..etc times of a rxn you multiply the rxn by that number as well as its KJ!

hope this helps!!