Answer: A. -396 kJ
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
[tex]2SO_2(g)+O_2 (g)\rightarrow 2SO_3(g)[/tex] [tex]\Delta H^0=198kJ[/tex]
Reversing the reaction, changes the sign of [tex]\Delta H[/tex]
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2 (g)[/tex] [tex]\Delta H^0=-198kJ[/tex]
On multiplying the reaction by 2, enthalpy gets multiplied by 2:
[tex]4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)[/tex] [tex]\Delta H=2\times -198kJ=-396kJ[/tex]
Thus the enthalpy change for the reaction [tex]4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)[/tex] is -396 kJ.
