Am I doing this correctly? (enthalpy change)?
A. S (s) + O2 (g) --> SO2 (g) delta H= -297kJ
B. 2S (s) + 3O2 (g) --> 2SO3 (l) delta H= -790kJ
Third equation:
C. 2SO2 (g) + O2 (g) -->2SO3 (l)
Do I just add the first two equations together? Or do I flip the second equation then add with the first equation? If so, is it right if I changed the sign of the second equation then add to do the first one?
Answer:
You need to flip the first chemical equation, which changes the sign on the enthalpy value from -297 to +297 (because a reaction that is exothemic in one direction must be endothermic in the opposite direction). However, you also have to multiply equation one by two to give:
2SO2 === 2S + 2O2
You also have to multiply the enthalpy by two; hence, it is now equal to (2)(+297). Now you can simply add the equations and see that the overall equation is now equ (3). To find the overall enthalpy add enthalpy change from reaction two to enthalpy for the reverse of reaction one; therefore, enthalpy change for three is equal to (2)(+297) + (-790). Now ask yourself: is the reaction exothermic or endothermic?
it looks like A is reverse and doubled so u reverse the Delta H and double delta H of "A" then add a and b therefore
-790+2*297 to get ur ans
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