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Sagot :
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 [tex]Delta H^o_{rxn}[/tex]
[tex]\Delta H^o_{rxn}=\sum (n \times \Delta H^o_{products})-\sum (n \times \Delta H^o_{reactants})[/tex] .....(1)
For the given chemical reaction:
[tex]3MnO_2(s)+4Al(s)\rightarrow 2Al_2O_3(s)+3Mn(s)[/tex]
The expression for the enthalpy change of the reaction will be:
[tex]\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))})][/tex]
Taking the standard heat of formation values:
[tex]\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[/tex]
Plugging values in the above expression:
[tex]\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[/tex]
Hence, the enthalpy of the reaction is -1791.31 kJ.
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