Question

3 upper M n upper O subscript 2 (s) plus 4 upper A l (s) right arrow 2 upper A l subscript 2 upper O subscript 3 (g) plus 3 upper M n (s). What is the enthalpy of the reaction?

Answer 1

Answer: The enthalpy of the reaction is -1791.31 kJ.

Explanation:

Enthalpy change is the difference between the enthalpies of products and the enthalpies of reactants each multiplied by its stoichiometric coefficients. It is represented by the symbol $Delta H^o_{rxn}$

$\Delta H^o_{rxn}=\sum (n \times \Delta H^o_{products})-\sum (n \times \Delta H^o_{reactants})$        .....(1)

For the given chemical reaction:

$3MnO_2(s)+4Al(s)\rightarrow 2Al_2O_3(s)+3Mn(s)$

The expression for the enthalpy change of the reaction will be:

$\Delta H^o_{rxn}=[(2 \times \Delta H^o_f_{(Al_2O_3(s))}) + (3 \times \Delta H^o_f_{(Mn(s))})] - [(3 \times \Delta H^o_f_{(MnO_2(s))}) + (4 \times \Delta H^o_f_{(Al(s))})]$

Taking the standard heat of formation values:

$\Delta H^o_f_{(Al_2O_3(s))}=-1675.7kJ/mol\\\Delta H^o_f_{(Al(s))}=0kJ/mol\\\Delta H^o_f_{(MnO_2(s))}=-520.03kJ/mol\\\Delta H^o_f_{(Mn(s))}=0kJ/mol$

Plugging values in the above expression:

$\Delta H^o_{rxn}=[(2 \times (-1675.7))+(3 \times 0)] - [(3 \times (-520.03))+(4 \times 0)]\\\\\Delta H^o_{rxn}=-1791.31 kJ$

Hence, the enthalpy of the reaction is -1791.31 kJ.