Question

Sulfur reacts with oxygen to form sulfur dioxide (SO2(g), Hf = –296.8 kJ/mol) according to the equation below.

Answer 1

Given that the reactants are the very consituents (elements)  of the final compound, the enthalpy change of this reaction is the same  heat of formation, Hf.

That means that the enthalpy chage for the reaction is -296.8 kj/mol

Answer 2

Answer : The enthalpy change for the reaction is -296.8 kJ/mol

Explanation :

Enthalpy change : It 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:  

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

The equilibrium reaction follows:

$S(s)+O_2(g)\rightleftharpoons SO_2(g)$

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

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(SO_2(g))})-(1\times \Delta H^o_f_{(S(s))}+1\times \Delta H^o_f_{(O_2(g))})$

We are given:

$\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(S(s))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=(1\times -296.8)-(1\times 0+1\times 0)=-296.8kJ/mol$

Therefore, the enthalpy change for the reaction is -296.8 kJ/mol