Respuesta :
The enthalpy change for the reaction is -296.8 kJ.
Explanation:
S(s) + O₂(g) ⇄ SO₂ (g)
We have to find the enthalpy change by finding the difference between the sum of enthalpies of products and that of reactants along with their coefficient terms. It is termed as ΔH.
It is found that,
[tex]\Delta H_{f^{\circ}}\left(S O_{2(g)}\right)=-296.8 \mathrm{kJ} / \mathrm{mol}[/tex]
[tex]\Delta H_{f^{\circ}}\left(S_{(s)}\right)=0 k J / m o l[/tex]
[tex]\Delta H_{f^{\circ}}\left(O_{2} _{(g)}\right)=0 k J / m o l[/tex]
The enthalpy change can be found out as,
[tex]\Delta H_{r x n}=\left(1 \times \Delta H_{f}\left(S O_{2}\right)_{(g)}\right)-\left(1 \times \Delta H_{f}(S)_{(s)}+1 \times \Delta H_{f}\left(O_{2}\right)(g)\right)[/tex]
= (1 × -296.8) - ((1 × 0) + ( 1 × 0 ))
= -296.8 kJ
So the enthalpy change for the reaction is -296.8 kJ/ mol.
