# Find the ΔH° for the reaction? a) Fe2O3(s)+3CO2(g)→2Fe(s)+3CO2(g)

##### 1 Answer
Aug 12, 2017

For

${\text{Fe"_2"O"_3(s) + 3"CO"(g) -> 2"Fe"(s) + 3"CO}}_{2} \left(g\right)$

you should get $- \text{23.47 kJ/mol}$, if you use my numbers. Go look up numbers from your own book and try again... see what you get, and verify mine is correct.

Well, when you manage to search your appendix for the substances in...

${\text{Fe"_2"O"_3(s) + 3"CO"(g) -> 2"Fe"(s) + 3"CO}}_{2} \left(g\right)$

...and you manage to fix your reaction, you should find them. I'm using NIST, which is quite a reliable database, especially for thermochemistry and spectroscopic data.

$\Delta {H}_{f , F {e}_{2} {O}_{3} \left(s\right)}^{\circ} = - \text{825.50 kJ/mol}$
http://webbook.nist.gov/cgi/cbook.cgi?ID=C1317608&Mask=2

$\Delta {H}_{f , C O \left(g\right)}^{\circ} = - \text{110.53 kJ/mol}$
http://webbook.nist.gov/cgi/cbook.cgi?ID=C630080&Units=SI&Mask=1

$\Delta {H}_{f , C {O}_{2} \left(g\right)}^{\circ} = - \text{393.52 kJ/mol}$
http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Mask=1

(Why don't I need to look up $\Delta {H}_{f}^{\circ}$ for $\text{Fe} \left(s\right)$?)

The standard enthalpy of reaction is then given by:

$\Delta {H}_{r x n}^{\circ} = {\sum}_{P} {\nu}_{P} \Delta {H}_{f , P}^{\circ} - {\sum}_{R} {\nu}_{R} \Delta {H}_{f , R}^{\circ}$

where:

• $\nu$ is the stoichiometric coefficient for product $P$ or reactant $R$.
• $\Delta {H}_{f}^{\circ}$ is the standard enthalpy of formation for a given substance in a given phase.

Thus, we get:

$\textcolor{b l u e}{\Delta {H}_{r x n}^{\circ}} = \left[2 \Delta {H}_{f , F e \left(s\right)}^{\circ} + 3 \Delta {H}_{f , C {O}_{2} \left(g\right)}^{\circ}\right] - \left[1 \Delta {H}_{f , F {e}_{2} {O}_{3} \left(s\right)}^{\circ} + 3 \Delta {H}_{f , C O \left(g\right)}^{\circ}\right]$

= overbrace([underbrace(2(0))_"2Fe(s)" + underbrace(3(-393.52))_("3CO"_2(g))])^"Products" - overbrace([underbrace(1(-825.50))_("Fe"_2"O"_3(s)) + underbrace(3(-110.53))_(3"CO"(g))])^"Reactants" $\text{kJ/mol}$

$= \left(0 - 1180.56 + 825.50 + 331.59\right)$ $\text{kJ/mol}$

$= \underline{\textcolor{b l u e}{- \text{23.47 kJ/mol}}}$