how in the world do you do this problem?
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
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!!
To find the enthalpy (ΔH) for the given reaction, you can use the concept of Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual reactions.
Here's how you can use Hess's Law to find the enthalpy change of the reaction:
1. First, write down the given thermochemical equations:
a) H2SO3(l) → H2O(l) + SO2(g) ΔH = 62 kJ
b) SO2(g) → S(s) + O2(g) ΔH = 297 kJ
c) H2S(g) + 1/2O2(g) → S(s) + H2O(l) ΔH = -155 kJ
2. Manipulate the equations to match the desired reaction:
a) Multiply the first equation by 2 to balance the number of oxygen atoms:
2H2SO3(l) → 2H2O(l) + 2SO2(g) ΔH = 124 kJ
b) Flip the second equation to match the production of SO2 in the first equation:
S(s) + O2(g) → SO2(g) ΔH = -297 kJ
c) Multiply the third equation by 2 to match the formation of H2S in the first equation:
2H2S(g) + O2(g) → 2S(s) + 2H2O(l) ΔH = -310 kJ
3. Add up the modified equations:
2H2SO3(l) + S(s) + O2(g) + 2H2S(g) + O2(g) → 2H2O(l) + 2SO2(g) + 2S(s) + 2H2O(l)
ΔH(total) = ΔH1 + ΔH2 + ΔH3
ΔH(total) = 124 kJ + (-297 kJ) + (-310 kJ)
ΔH(total) = -483 kJ
Therefore, the enthalpy change (ΔH) for the reaction:
H2SO3(l) → H2S(g) + 3/2O2(g) is -483 kJ.