Chapter 4, Problem 4.28NP

Interpretation Introduction

Interpretation:

From the data given at 298.15 K, the standard enthalpy of formation of FeO(s) and Fe₂O₃(s) needs to be calculated.

	$\Delta H_R^o\left(\text{kJ/mol}\right)$
${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3\text{C}\left(\text{graphite}\right)\to 2\text{Fe}\left(s\right)+3\text{CO}\left(g\right)$	492.6
$\text{FeO}\left(s\right)+\text{C}\left(\text{graphite}\right)\to \text{Fe}\left(s\right)+\text{CO}\left(g\right)$	155.8
$\text{C}\left(\text{graphite}\right)+{\text{O}}_{\text{2}}\to {\text{CO}}_2\left(g\right)$	-393.51
$\text{CO}\left(g\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)$	-282.98

Concept Introduction :

Enthalpy is known as the sum of internal energy and product of pressure and volume provided by the below equation:

  $H=U+PV$ .

When the procedure happens at constant pressure, the evolved heat is equal to the change in enthalpy.