The enthalpy change for the reaction of titanium metal with gaseous iodine is given by the following thermochemical equation: 2 Ti(s) + 3 I2(g) → 2 TiI3(s) ΔH rxn = −839 kJ What is the enthalpy change for the reaction below? TiI3(s) → Ti(s) + 3/2 I2(g)

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.

The given chemical reaction is,

[tex]2Ti(s)+3I_2(g)\rightarrow 2TiI_3(s)[/tex] [tex]\Delta H=-839kJ[/tex]

Now we have to determine the enthalpy change for the reaction below:

[tex]TiI_3(s)\rightarrow Ti(s)+\frac{3}{2}I_2(g)[/tex] [tex]\Delta H'=?[/tex]

By reversing and then dividing the reaction by 2, we get the enthalpy change for the reaction.

The expression will be:

[tex]\Delta H'=-\frac{(\Delta H)}{2}[/tex]

[tex]\Delta H'=-\frac{(-839kJ)}{2}[/tex]

[tex]\Delta H'=419.5kJ[/tex]

Therefore, the enthalpy change for the reaction is, 419.5 kJ