Hi, please help with this question. Thanks.
Find the ΔH for the reaction 1/2 H2(g) + 1/2 Cl2(g) -> HCl(g), given the following reactions and subsequent ΔH values:
COCl2(g) + H2O(l) -> CH2Cl2(l) + O2(g) ΔH= 47.5 kJ
2HCl(g) + 1/2 O2(g) -> H2O(l) + Cl2(g) ΔH= 105 kJ
CH2Cl2(l) + H2(g) + 3/2 O2(g) -> COCl2(g) + 2H2O(l) ΔH= -402.kJ
Chemistry help needed badly?
2017-07-06 9:09 pm
回答 (2)
2017-07-06 10:29 pm
✔ 最佳答案
Rewrite the three given thermochemical equations as :(1/2)COCl₂(g) + (1/2)H₂O(l) → (1/2)CH₂Cl₂(l) + (1/2)O₂(g) …. ΔH = (1/2)(+47.5) kJ = +23.8 kJ
(1/2)H₂O(l) + (1/2)Cl₂(g) → HCl(g) + (1/4)O2(g) …. ΔH= (-1/2)(105) kJ = -52.5 kJ
(1/2)CH₂Cl₂(l) + (1/2)H₂(g) + (3/4)O₂(g) → (1/2)COCl₂(g) + H₂O(l) …. ΔH = (1/2)(-402) kJ = -201 kJ
Add the above three thermochemical equations, and cancel (1/2)COCl₂(g), H₂O(l), (1/2)CH₂Cl₂(l) and (3/4)O2(g) on the both sides.
(1/2)H₂(g) + (1/2)Cl₂(g) → HCl(g) …. ΔH = (+23.8) + (-52.5) + (-201) kJ = -230 kJ (to 3 sig. fig.)
2017-07-06 9:30 pm
you have to manipulate the given equations such that when you add them you get the equation you want. (a Hess's law problem)
so 1/2 times equation 3, plus -1/2 times equation 2, plus 1/2 times equation 1 should do it.
so 1/2 times equation 3, plus -1/2 times equation 2, plus 1/2 times equation 1 should do it.
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