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

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:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$SO_2(g)+\frac{1}{2}O_2(g)\rightarrow SO_3(g)$

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

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

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:

$\Delta H^o_{rxn}=[(1\times -395.7)]-[(1\times (-296.84))+(\frac{1}{2}\times 0)]\\\\\Delta H^o_{rxn}=-98.86kJ/mol$

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