Question

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.

Answer 1

Answer: The change in enthalpy of the reaction is -26.45 kJ

Explanation:

The given chemical reaction follows:

$H_2(g)+I_2(s)\rightarrow 2HI(g);\Delta H=52.9kJ$

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:

$HI(g)\rightarrow \frac{1}{2}H_2(g)+\frac{1}{2}I_2(s)$

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 = $\frac{-52.9}{2}\times 1=-26.45kJ$

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