Given the following reaction: h2(g)+i2(s) → 2hi(g) with a ∆h of 52.9 kj. what is the change in enthalpy for the following reaction: hi(g) → 1 2h2(g)+1 2 i2(s)? express your answer in kj.

The change in enthalpy of the reaction is -26.45 kJ

Explanation:

The given chemical reaction follows:

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:

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 =

Hence, the change in enthalpy of the reaction is -26.45 kJ