Question

Based on the information above, what is ΔH° for the reaction SO2(g)+12O2(g)→SO3(g) ?

Answer 1

Answer: The standard enthalpy change of the given reaction is -98.86 kJ/mol

Step-by-step explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as
`\Delta H^o`

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H^o_f_((product))]-\sum [n* \Delta H^o_f_((reactant))]`

For the given chemical reaction:

`SO_2(g)+(1)/(2)O_2(g)\rightarrow SO_3(g)`

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

`\Delta H^o_(rxn)=[(1* \Delta H^o_f_((SO_3(g))))]-[(1* \Delta H^o_f_((SO_2(g))))+((1)/(2)* \Delta H^o_f_((O_2(g))))]`

We are given:

`\Delta H^o_f_((SO_3(g)))=-395.7kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol\\\Delta H^o_f_((SO_2(g)))=-296.84kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(1* -395.7)]-[(1* (-296.84))+((1)/(2)* 0)]\\\\\Delta H^o_(rxn)=-98.86kJ/mol`

Hence, the standard enthalpy change of the given reaction is -98.86 kJ/mol