Question

Rxn

Answer 1

Answer: The enthalpy of formation of $SO_3$ is  -396 kJ/mol

Explanation:

Calculating the enthalpy of formation of $SO_3$

The chemical equation for the combustion of propane follows:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

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

$\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ$

Putting values in above equation, we get:

$-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol$

The enthalpy of formation of $SO_3$ is -396 kJ/mol