Answer to Question #161211 in General Chemistry for neyola rebello

Question

Given the following data:

Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g)      ΔH = -26.8 kJ

   FeO(s) + CO(g) → Fe(s) + CO₂(g)             ΔH = -16.5 kJ

Calculate the ΔH for the reaction below:

Fe₂O₃(s) + CO(g) → 2 FeO(s) + CO₂(g)

Answer 1

Solution:

Given the chemical reaction:

Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g)      ΔH = -26.8 kJ

FeO(s) + CO(g) → Fe(s) + CO₂(g)             ΔH = -16.5 kJ

Firstly we should find the standard enthalpies of Fe₂O₃(s) and FeO(s) in terms of Hess`s law:

$\Delta H_{reaction}=\Delta H_{product}-\Delta H_{reactant}$

$\Delta H(Fe_2O_3) = -822.2 (\frac{kJ}{mole})$

$\Delta H(FeO)=-264.8 (\frac{kJ}{mole})$

Then, we put these values to the main chemical reaction:

$Fe_2O_3+CO\rightarrow2FeO+CO_2$

Hess`s law:

$\Delta H_{reaction}=\Delta H_{product}-\Delta H_{reactant}$

$\Delta H_{product}=$ 2(-264.8 $(\frac{kJ}{mole})$ ) + (-393.5 $(\frac{kJ}{mole})$ )= -923.1 kJ

$\Delta H_{reactant}=$ (-822.2 $(\frac{kJ}{mole})$ ) + (-110.5 $(\frac{kJ}{mole})$ ) = -932.7 kJ

$\Delta H_{reaction}=$ (-923.1 kJ) - ( -932.7 kJ) = 9.6 kJ

Answer:

$\Delta H_{reaction}=9.6kJ$