Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g)

Use the following:
H2 (g) + 1/2O2 (g) → H2O (l) ΔH = -285.9 kJ
H2O (s) → H2O (l) ΔH = +6.0 kJ

A. –291.9 kJ
B. +291.9 kJ
C. –279.9 k
D. +279.9 kJ

Question

Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g)

Use the following:
 H2 (g) + 1/2O2 (g) → H2O (l)  ΔH = -285.9 kJ
H2O (s) → H2O (l)                      ΔH = +6.0 kJ

 
	A. –291.9 kJ	
	B. +291.9 kJ	
 	C. –279.9 k	
	D. +279.9 kJ

travisbrown372 · Answer

heres examples https://www.chem.wisc.edu/areas/clc/general/104/practprob_3_104-2.pdf

wolfe8 · Answer

From what I know, you are basically reversing the first given reaction(change sign) but include the difference in the state of oxygen. So you add the second energy change but note the change of state. Does that seem right to you or make sense?

anonymous · Answer

Yeah it makes sense, I just didn't know if it was possible to have 2 positive kJ in this

wolfe8 · Answer

Some reactions produce more energy than they absorb. They are called exothermic reactions. I'm not sure if this is what you were wondering.