FeS2(s) = −178.2 kJ/mol FeCl2(s) = −341.8 kJ/mol
FeCl3(s) = −399.5 kJ/mol HCl(g) = −92.3 kJ/mol
HCl(aq) = −167.1 kJ/mol H2S(g) = −20.6 kJ/mol
H2O(g) = −271.8 kJ/mol
Calculate the H° for the reaction below.
2FeS2(s) + 8HCl(g) 2FeCl3(s) + 4H2S(g) + Cl2(g)
The standard enthalpies of formation for several substances are given below:?
2020-09-02 3:56 pm
回答 (2)
2020-09-02 5:04 pm
✔ 最佳答案
The given data of enthalpies of formation for FeCl₂(s), HCl(g) and H₂O(g) are redundant.2FeSₛ(s) + 8HCl(g) → 2FeCl₃(s) + 4H₂S(g) + Cl₂(g) ΔH° = ?
Enthalpy of reaction, ΔH°
= 2 ΔH°f[FeCl₃(s) + 4 ΔH°f[H₂S(g) + ΔH°f[Cl₂(g)] - 2 ΔH°f[FeS₂(s)] - 8 ΔH°f[HCl(g)]
= [2(-399.5) + 4(-20.6) + (0) - 2(-178.2) - 8(-167.1)] kJ
= +811.8 kJ
2020-09-02 4:05 pm
It means nothing at all. The answer would be FeS2S is 175. Yours is wrong
收錄日期: 2021-04-24 07:58:43
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