Answer : The enthalpy change for this reaction is -193.8 kJ.
Solution :
The balanced chemical reaction is,
[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{SO_3}\times \Delta H_{SO_3})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{SO_2}\times \Delta H_{SO_2})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]\Delta H=[(2\times -395.7)]-[(1\times 0)+(2\times -298.8)][/tex]
[tex]Delta H=-193.8kJ[/tex]
Therefore, the enthalpy change for this reaction is, -193.8 KJ
