Help with enthalpy problem?

[匿名]

2018-06-13 8:49 am

A scientist measures the standard enthalpy change for the following reaction to be 186.8 kJ :

Fe3O4(s) + 4 H2(g) ----> 3Fe(s) + 4 H2O(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of Fe3O4(s) is _____ kj/mol

A scientist measures the standard enthalpy change for the following reaction to be 154.3 kJ :

CaCO3(s) ----> CaO(s) + CO2(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CaO(s) is

回答 (1)

wanszeto

2018-06-13 9:56 am

✔ 最佳答案

Refer to: http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf
The standard enthalpy changes of formation :
ΔHf°[H₂O(g)] = -241.8 kJ/mol
ΔHf°[CaCO₃(g)] = -1207.0 kJ/mol
ΔHf°[CO₂(g)] = -393.5 kJ/mol

The standard enthalpy change of any element is zero.
ΔHf°[H₂(g)] = 0 kJ/mol
ΔHf°[Fe(g)] = 0 kJ/mol

====
1.
Fe₃O₄(s) + 4 H₂(g) → 3Fe(s) + 4 H₂O(g) …… ΔHrnx° = +186.8 kJ

ΔHrnx° = 3 ΔHf°[Fe(s)] + 4 ΔHf°[H₂O(g)] - ΔHf°[Fe₃O₄(s)] - 4 ΔHf°[H₂(g)]
+186.8 = 3(0) + 4(-241.8) - ΔHf°[Fe₃O₄(s)] - 4(0)
ΔHf°[Fe₃O₄(s)] = 4(-241.8) - 186.8 kJ/mol
ΔHf°[Fe₃O₄(s)] = -1154.0 kJ/mol

====
2.
CaCO₃(s) → CaO(s) + CO₂(g) …… ΔHrxn° = +154.3 kJ

ΔHrxn° = ΔHf°[CaO(s)] + ΔHf°[CO₂(g)] - ΔHf°[CaCO₃(s)]
+154.3 = ΔHf°[CaO(s)] + (-393.5) - (-1207.0)
ΔHf°[CaO(s)] = 154.3 - (-393.5) + (-1207.0) kJ/mol
ΔHf°[CaO(s)] = -659.2 kJ/mol

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