Question

Question : Given the following data:

${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{2}\mathbf{Fe}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$ ∆H = -23 kJ

$\mathbf{3}{\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(s\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{CO}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{2}{\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\left(s\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)$ ∆H = -39 kJ

${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\left(s\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{CO}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{3}\mathbf{FeO}\left(s\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)$ ∆H = 18 kJ

Calculate ∆H for the reaction

$\mathbf{FeO}\left(s\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{CO}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}\mathbf{Fe}\left(s\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)$

Short answer

∆H for the reaction = -11 kJ

Step-by-step solution

Multiply equation 1 by 3

3${\mathrm{Fe}}_2{\mathrm{O}}_3$ + 9CO $\to$6Fe + 9CO₂ ∆H = -23×3 = -69 kJ

Reverse equation 2

$2{\mathrm{Fe}}_3{\mathrm{O}}_4+{\mathrm{CO}}_2\to 3{\mathrm{Fe}}_2{\mathrm{O}}_3+\mathrm{CO}$ ∆H = 39 kJ

Reverse equation 3 and multiply by 2

$6\mathrm{FeO}+2{\mathrm{CO}}_{2}2{\mathrm{Fe}}_3{\mathrm{O}}_4+2\mathrm{CO}$ ∆H = -18×2 = -36 kJ

Write all equations and solve them by the Hess law

3${\mathrm{Fe}}_2{\mathrm{O}}_3$+ 9CO $\to$6Fe + 9${\mathrm{CO}}_2$ ∆H = -69 kJ

2${\mathrm{Fe}}_3{\mathrm{O}}_4$ + C${\mathrm{O}}_2$$\to$3${\mathrm{Fe}}_2{\mathrm{O}}_3$ + CO ∆H = 39 kJ

6FeO + 2${\mathrm{Fe}}_2{\mathrm{O}}_3$$\to$ 2role="math" localid="1648667451571" ${\mathrm{Fe}}_3{\mathrm{O}}_4$+ 2CO ∆H = -36 kJ

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6FeO+ 6CO $\to$6 Fe + 6${\mathrm{CO}}_2$ ∆H = -66 kJ

Divide this equation by 6:

role="math" localid="1648668265488" $\mathrm{FeO}+\mathrm{CO}\to \mathrm{Fe}+{\mathrm{CO}}_2$ ∆H = -11 kJ