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

Use the following:

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


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asked by Angela
  1. If we look at this the reverse direction.
    H2(g) + 1/2O2(g) --> H2O(l) that's the heat formation liquid H2O in kJ/mol and my text shows that as -285.8 kJ/mol
    Then H2O(l) ==> H2O(s) is another 6.0 kJ/mol to form ice and freezing water to ice is exothermic so that should be -6.0 kJ/mol. So I would go with -285.8 - 6 = ? Now you want the reverse of that so change the sign.

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