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leva [86]
4 years ago
7

Rxn

Chemistry
1 answer:
givi [52]4 years ago
4 0

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

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