Asked by Emily
Find the heat of reaction (ΔH) for each of the following chemical reactions and note whether each reaction is exothermic or endothermic.
1. H2O(l) -> H2O(g)
This is what I have so far:
H2(g) + ½ O2(g) -> H2O(l) ΔH = -286.0 kJ/mol
H2(g) + ½ O2(g) -> H2O(g) ΔH = -242.0 kJ/mol
H2O(l) -> H2(g) + ½ O2(g) ΔH = +286.0 kJ/mol
1 mol x H2(g) + ½ O2(g) -> H2O(g) ΔH = 1 mol x (-242.0 kJ/mol)
1 mol x H2O(l) -> H2(g) + ½ O2(g) ΔH = 1 mol x (+286.0 kJ/mol)
H2(g) + ½ O2(g) + H2O(l) -> H2O(g) + H2(g) + ½ O2(g) ΔH = 44 kJ
H2O(l) -> H2O(g) ΔH = 44 kJ
Am I on the right track?
1. H2O(l) -> H2O(g)
This is what I have so far:
H2(g) + ½ O2(g) -> H2O(l) ΔH = -286.0 kJ/mol
H2(g) + ½ O2(g) -> H2O(g) ΔH = -242.0 kJ/mol
H2O(l) -> H2(g) + ½ O2(g) ΔH = +286.0 kJ/mol
1 mol x H2(g) + ½ O2(g) -> H2O(g) ΔH = 1 mol x (-242.0 kJ/mol)
1 mol x H2O(l) -> H2(g) + ½ O2(g) ΔH = 1 mol x (+286.0 kJ/mol)
H2(g) + ½ O2(g) + H2O(l) -> H2O(g) + H2(g) + ½ O2(g) ΔH = 44 kJ
H2O(l) -> H2O(g) ΔH = 44 kJ
Am I on the right track?
Answers
Answered by
DrBob222
Yes but I don't understand why you did all of that work.
dH = (n*dHf products) - (n*dHf reactants)
dH = (1*-242 kJ) - (1*-286 kJ)
dH = -242-(-286) = 44 kJ.
Since dH is +, the reaction is endothermic. I looked up the H<sup>o</sup><sub>f</sub> in a set of tables.
dH = (n*dHf products) - (n*dHf reactants)
dH = (1*-242 kJ) - (1*-286 kJ)
dH = -242-(-286) = 44 kJ.
Since dH is +, the reaction is endothermic. I looked up the H<sup>o</sup><sub>f</sub> in a set of tables.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.