Question

Enter your answer in the provided box.S(rhombic) + O2(g) → SO2(g) ΔHo rxn= −296.06 kJ/molS(monoclinic) + O2(g) → SO2(g) ΔHo rxn= −296.36 kJ/molcalculate the enthalpy change for the transformationS(rhombic) → S(monoclinic)(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)_______kJ/mol

Answer 1

Answer: $\Delta H^0=+0.3kJ/mol$.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$S_{rhombic}+O_2(g)\rightarrow SO_2(g)$    $\Delta H^0_1=-296.06kJ$   (1)

$S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ$  (2)

The final reaction is:  

$S_{rhombic}\rightarrow S_{monoclinic}$  $\Delta H^0_3=?$   (3)

By subtracting (1) and (2)

$\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ$

Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ