Asked by anonymous
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
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
Answers
Answered by
DrBob222
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.
Answered by
anonymous
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.
Answered by
DrBob222
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)
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)
Answered by
DrBob222
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.)