Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
[tex]\Delta (H_{f})_{MnO_{2}}[/tex] = -520.0 KJ/mole
[tex]\Delta (H_{f})_{Al_{2}O_{3}}[/tex] = -1699.8 KJ/mole
The balanced chemical reaction is,
[tex]3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)[/tex]
Formula used :
[tex]\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}[/tex]
[tex]\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))[/tex]
We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
[tex]\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})][/tex]
= (-3399.6) + (1560)
= -1839.6 KJ
