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
so you just add 62kj +297kg+155kj
to get 514. That seems easy enough
Your target equation is not apparent from your post; however, if the target equation is the first one in the post, adding the other three will not get the equation you want.
well can you help me. I have to do other problems similar to this and i've been looking it up in my book and online and cant seem to get the hang of how to do it.
Label the equation you have as 1, 2, 3.
Use equation 1 as is.
Use equation 2 as is.
Reverse equation 3 (change the sign of DH when you reverse an equation), than add all of them and the new DHs. Check to make sure the equation is what you wnat, the It will be DH(1) + DH(2) + (-DH)(3)
That last line I wrote could be confusing. You add the three NEW equations as outlined above. Then you add the three NEW DHs (the equation 1 DH as is, the equation 2 DH as is, and the NEW DH for equation 3.)
To determine the enthalpy (ΔH) for the reaction: H2SO3(l) → H2S(g) + 3/2O2(g), you can use Hess's Law, which states that the enthalpy change in a reaction is independent of the pathway taken.
1. Begin by writing the given thermochemical equations and their corresponding enthalpy values:
H2SO3(l) → H2O(l) + SO2(g) ΔH = +62 kJ
SO2(g) → S(s) + O2(g) ΔH = +297 kJ
H2S(g) + 1/2O2(g) → S(s) + H2O(l) ΔH = -155 kJ
2. Notice that the intermediate product of H2O(l) is present in both the first and third equations. This means we can cancel out the H2O(l) from those equations by manipulating their coefficients and reversing the direction:
H2SO3(l) → H2O(l) + SO2(g) ΔH = +62 kJ
2H2O(l) + 2SO2(g) → 2H2SO3(l) ΔH = -62 kJ
H2S(g) + 1/2O2(g) → S(s) + H2O(l) ΔH = -155 kJ
2H2O(l) + 2S(s) + O2(g) → 2H2S(g) ΔH = +155 kJ
3. Now, add up the manipulated equations to obtain the overall balanced equation for the reaction:
2H2SO3(l) + 2H2O(l) + 2SO2(g) + O2(g) → 2H2S(g) + 3O2(g) + S(s)
4. Finally, sum up the corresponding enthalpy values for each equation:
ΔH1 + ΔH2 + ΔH3 = (+62 kJ) + (+297 kJ) + (-155 kJ) = +204 kJ
Therefore, the enthalpy change for the reaction H2SO3(l) → H2S(g) + 3/2O2(g) is +204 kJ.