Question

The following data are needed for this question.
Hf (CO(g)) = –111 kJ mol–1
Hf(CO2(g)) = –394 kJmol–1
Hf(Fe2O3(s)) = –822 kJ mol–1
Carbon monoxide reacts with iron(III) oxide.
3CO(g) + Fe2O3(s) → 3CO2(g) + 2Fe(s)
What is the enthalpy change when 55.8 g of iron are produced by this reaction?
A. –27.0 kJ
B. –13.5 kJ
C. +13.5 kJ
D. +27.0 kJ

Answer 1

Answer: The correct answer is Option B.

Explanation:

For the given chemical reaction:

$3CO(g)+Fe_2O_3(s)\rightarrow 3CO_2(g)+2Fe(s)$

The equation used to calculate enthalpy change of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(Fe(s))})]-[(3\times \Delta H^o_f_{(CO(g))})+(1\times \Delta H^o_f_{(Fe_2O_3(s))})]$

We are given:

$\Delta H^o_f_{(CO_2(g))}=-394kJ/mol\\\Delta H^o_f_{(Fe(s))}=0kJ/mol\\\Delta H^o_f_{(CO(g))}=-111kJ/mol\\\Delta H^o_f_{(Fe_2O_3(s))}=-822kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(3\times (-394))+(2\times (0))]-[(3\times (-111))+(1\times (-822))]=-27kJ/mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of iron = 55.8 g

Molar mass of iron = 55.8 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{55.8g}{55.8g/mol}=1mol$

By Stoichiometry of the reaction:

When 2 moles of iron is produced, the amount of heat released is 27 kJ

So, when 1 mole of iron is produced, the amount of heat released will be = $\frac{27}{2}\times 1=13.5kJ$

Hence, the correct answer is Option B.