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Calculate the standard enthalpy of the reaction. - Chemistry

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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
Numerical

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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Chapter 4: Chemical Thermodynamics - Long answer questions

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Chapter 4 Chemical Thermodynamics
Long answer questions | Q 2.3

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