Answer: The change in enthalpy of the reaction is -26.45 kJ
Explanation:
The given chemical reaction follows:
[tex]H_2(g)+I_2(s)\rightarrow 2HI(g);\Delta H=52.9kJ[/tex]
The enthalpy change of the reaction when 2 moles of HI is formed is 52.9 kJ
Now, the chemical reaction for which enthalpy change is to be calculated is:
[tex]HI(g)\rightarrow \frac{1}{2}H_2(g)+\frac{1}{2}I_2(s)[/tex]
Sign for enthalpy change will be reversed, because the reaction is reverse.
By Stoichiometry of the reaction:
When 2 moles of Hi is reacted, the enthalpy change of the reaction is -52.9 kJ
So, when 1 mole of HI will be reacted, the enthalpy change of the reaction will be = [tex]\frac{-52.9}{2}\times 1=-26.45kJ[/tex]
Hence, the change in enthalpy of the reaction is -26.45 kJ
