Calculate the delta H for S + O2 = SO2

Question

Given the following data:

S + 3/2 O2 = SO3 delta H=-395.2kJ

2SO2 + O2 = 2SO3 delta H= -198.2kJ

Answer 1

Find an answer to your problem here.

https://youtu.be/qv-EwEGgGBM

Answer 2

Hi Callyn W.,

You want to add your equations so that you get your intended final equation, right?

There are very many routes to accomplish the same thing; here's one of them. Look at your desired equation: S + O2 = SO2

It has the S on the left.

So pick an equation from your input list with an S on the left:

S + (3/2)O2 = SO3

Now, are you lacking any reactants on the left? Not really. But you have that SO3 on the right! So, prepare to get rid of it by making it appear on the left in an equation, so as to cancel the one on the right when you add the 2 equations. How to do that?

Turn around your second input equation, and multiply it throughout by the factor (1/2):

(2SO3 = 2SO2 + O2)*0.5 --> SO3 = SO2 + (1/2)O2 associated with delta H = (+198.2)*0.5 . Why: the sign flips when you flip the equation, and the enthalpy number drops by 0.5 when you scale the reaction by 0.5.

Now you can write your equations and add them:

S + (3/2)O2 = SO3 .delta.H = -395.2kJ (note that we don't say "per mole" which would be ambigous, rather just leave it as is and know that it means for the equation as written)

SO3 = SO2 + (1/2) O2 .delta.H = +99.1 kJ

adding: S + O2 + SO2 .delta.H = (-395.2)+(99.1) kJ

Note that the O2 result on the left is the NET of the (3/2)O2 on the left, and the (1/2) O2 on the right, as addends. Why? Because if you HAD brought the (1/2)O2 over from the right to the left, it would have appeared there as a -(1/2) O2 -- the only way you could "MOVE IT" is to subtract it from both sides of the equation, just as you move anything around in an equation!

Hope this helped you. The key point is to start with at least one molecule you want, and then successively message the equations down until everything else falls into place (as it must!)

-- Cheers, -- Mr. d.