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 [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]S(s)+O_2(g)\rightleftharpoons SO_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\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))})[/tex]
We are given:
[tex]\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[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=(1\times -296.8)-(1\times 0+1\times 0)=-296.8kJ/mol[/tex]
Therefore, the enthalpy change for the reaction is -296.8 kJ/mol
