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.

Step-by-step 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 * \Delta H^o_(products))-\sum (n * \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 * \Delta H^o_f_((Al_2O_3(s)))) + (3 * \Delta H^o_f_((Mn(s))))] - [(3 *  \Delta H^o_f_((MnO_2(s)))) + (4 * \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 * (-1675.7))+(3 * 0)] - [(3 * (-520.03))+(4 * 0)]\\\\\Delta H^o_(rxn)=-1791.31 kJ`

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