Answer: The enthalpy of formation of [tex]SO_3[/tex] is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of [tex]SO_3[/tex]
The chemical equation for the combustion of propane follows:
[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ[/tex]
Putting values in above equation, we get:
[tex]-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol[/tex]
The enthalpy of formation of [tex]SO_3[/tex] is -396 kJ/mol
