[hamil]

Given: Hf in kJ/mol  (heat of formation)
H2O2 = 187.8
O2 = 0
H2O = 285.8
Calculate the enthalpy of reaction for decomposition of hydrogen peroxide into oxygen and water and determine if the reaction is exothermic or endothermic.

2H2O -> O2 + 2H2O

I calculated endothermic with heat of reaction 196 kJ

Can someone confirm or deny that I am correct.

[Corribus]

Well, it would be right, but I think your heats of formation are incorrect.  Heat of formation of water is -285.8 kJ/mol and I believe the sign for your hydrogen peroxide value is wrong as well. 

[hamil]

So, the correct equation should be..
2(-187. -> 1(0) + 2(-285) + X
-374 -> -570 + X
solving gives X = 196
This means the reaction is endothermic since 196 is on the right side of the equation, yes?

[Corribus]

No, the enthalpy change of the reaction, ΔH, is equal to

ΔH = 2*ΔH_f {water} + ΔH_f {dioxygen} - 2*ΔH_f {hydrogen peroxide}

(your intitial reaction equation also has a typo)

If ΔH_f {water} = -285.8 kJ/mol and ΔH_f {hydrogen peroxide} = -187.8 kJ/mol and ΔH_f {dioxygen} = 0, what does ΔH equal?

(Just a note - often you see this reaction written as H₂O₂ H₂O + 1/2 O₂.  This will impact your answer for the ΔH of the reaction.  But this is OK.  The answer you get for your reaction will be the enthalpy change for 2 moles of hydrogen peroxide decomposing.  The answer we would get for this one would be the enthalpy change for 1 mole of hydrogen peroxide decomposing.)

[hamil]

It looks like Delta H = -196.

Now, I am a bit confused by the signs. Is the reaction endo or exo. I thought it would be endothermic.

[Corribus]

Reaction is exothermic.  Assuming all the O-H bonds formed/broken have the same energy, the overall change in enthalpy comes from breaking a pretty weak oxygen-oxygen single bond (bond energy ~ 140 kJ/mol) and forming a pretty strong oxygen-oxygen pi-bond (~ 490 kJ/mol).  For every two O-O bonds broken, one O=O bond is formed.  Forming bonds releases energy, breaking bonds consumes energy.  You see the difference is about 200 kJ/mol released energy (140 kJ/mol*2 - 490 kJ/mol = -210 kJ/mol).  This crude approach is reasonably close to what you calculate from heats of formation.  So, this reaction is exothermic primarily due to the weakness of the O-O single bond, which just doesn't store very much energy compared to dioxygen.  The energy difference is released as heat. 

The reaction is also spontaneous because the entropy change is positive.  The primary reason for this is that you are forming a gas.

Therefore as you can imagine it is very favorable (large negative ΔG).  Despite this, it's pretty slow, but that's a topic for another day.