1. Tardigrade
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  4. On Fe2O3(.s.)+(3/2)C(.s.) arrow (3/2)CO2(.g.)+2Fe(.s.);Δ H° =+234.1kJ C(.s.)+O2(.g.) arrow CO2(.g.);Δ H° =-393.5 kJ Use these equations and Δ H° values to calculate Δ H° for this reaction 4Fe(.s.)+3O2(.g.) arrow 2Fe2O3(.s.)

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Q. On $Fe_{2}O_{3}\left(\right.s\left.\right)+\frac{3}{2}C\left(\right.s\left.\right) \rightarrow \frac{3}{2}CO_{2}\left(\right.g\left.\right)+2Fe\left(\right.s\left.\right);\Delta H^\circ =+234.1kJ$
$C\left(\right.s\left.\right)+O_{2}\left(\right.g\left.\right) \rightarrow CO_{2}\left(\right.g\left.\right);\Delta H^\circ =-393.5$ kJ
Use these equations and $\Delta H^\circ $ values to calculate $\Delta H^\circ $ for this reaction
$4Fe\left(\right.s\left.\right)+3O_{2}\left(\right.g\left.\right) \rightarrow 2Fe_{2}O_{3}\left(\right.s\left.\right)$

NTA AbhyasNTA Abhyas 2022

Solution:

$2Fe\left(\right.s\left.\right)+\frac{3}{2}CO_{2}\left(\right.g\left.\right) \rightarrow Fe_{2}O_{3}\left(\right.s\left.\right)+\frac{3}{2}C\left(\right.s\left.\right);\Delta H^\circ =-234.1kJ$ …(i)
$C\left(\right.s\left.\right)+O_{2}\left(\right.g\left.\right) \rightarrow CO_{2}\left(\right.g\left.\right)\left(\right.\Delta H^\circ =-393.5kJ\left.\right)$ …(ii)
Multiplying eq. (i) by 2 and eq. (ii) by 3 and on adding both equations, we get
$4Fe\left(\right.s\left.\right)+3O_{2}\left(\right.g\left.\right) \rightarrow 2Fe_{2}O_{3}\left(\right.s\left.\right);\Delta H^\circ =\left(\right.-234.1\times 2\left.\right)+\left(\right.-3\times 393.5\left.\right)$
$=-1648.7$ kJ