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Question
Calculate the standard enthalpy of the reaction.
\[\ce{2Fe_{(s)} + \frac{3}{2} O_{2(g)} -> Fe2O_{3(s)}}\]
Given:
| 1. | \[\ce{2Al_{(s)} + Fe2O_{3(s)} -> 2Fe_{(s)} + Al_2O_{3(s)}}\], | ∆rH° = –847.6 kJ |
| 2. | \[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\], | ∆rH° = –1670 kJ |
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Solution
Given: Given equations are,
\[\ce{2Al_{(s)} + Fe2O_{3(s)} -> 2Fe_{(s)} + Al_2O_{3(s)}}\], ∆rH° = –847.6 kJ .....(1)
\[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\], ∆rH° = –1670 kJ .......(2)
To find: Standard enthalpy of the given reaction
Calculation: Reverse equation (1),
\[\ce{2Fe_{(s)} + Al_2O_{3(s)} -> 2Al_{(s)} + Fe2O_{3(s)}}\], ΔrH° = 847.6 kJ …. (3)
Add equation (2) to equation (3),
| \[\ce{2Fe_{(s)} + Al_2O_{3(s)} -> 2Al_{(s)} + Fe2O_{3(s)}}\], | ΔrH° = 847.6 kJ |
| \[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\], | ∆rH° = –1670 kJ |
| \[\ce{2Fe_{(s)} + \frac{3}{2} O_{2(g)} -> Fe2O_{2(s)}}\], | |
ΔrH° = 847.6 + (−1670) = −822.4 kJ
The standard enthalpy of the given reaction is −822.4 kJ.
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