The standard enthalpies of formation of SO2 and SO3 are ‐297 and ‐396 kJ/mol respectively. Calculate the standard enthalpy of reaction for the reaction: SO2(g) + 1/2 O2(g) -->SO3 (g)

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H[/tex]

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

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

For the given chemical reaction:

[tex]SO_2(g)+\frac{1}{2}O_2(g)\rightarrow SO_3(g)[/tex]

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

[tex]\Delta H_{rxn}=[(1\times \Delta H_f_{(SO_3)})]-[(1\times \Delta H_f_{(SO_2)})+(\frac{1}{2}\times \Delta H_f_{(O_2)})][/tex]

We are given:

[tex]\Delta H_f_{(SO_3)}=-396kJ/mol\\\Delta H_f_{(SO_2)}=-297kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H_{rxn}=[(1\times (-396))]-[(1\times (-297))+(\frac{1}{2}\times (0))]\\\\\Delta H_{rxn}=-99kJ[/tex]

Hence, the enthalpy of the reaction is coming out to be -99 kJ.