Question

½ H₂(g) + ½ I₂(s) --> HI(g) ΔH= 26 kJ/mol

Answer 1

Answer : The correct option is, (C) 31 kJ/mole

Explanation :

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.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The sublimation of iodine reaction is,

$I_2(s)\rightarrow I_2(g)$    $\Delta H=?$

The intermediate balanced chemical reaction are,

(1) $\frac{1}{2}H_2(g)+\frac{1}{2}I_2(s)\rightarrow HI(g)$     $\Delta H_1=26kJ/mol$

(2) $\frac{1}{2}H_2(g)+\frac{1}{2}I_2(g)\rightarrow HI(g)$     $\Delta H_2=-5.0kJ/mol$

Now we are reversing reaction 2 and then adding both the equations, we get :

(1) $\frac{1}{2}H_2(g)+\frac{1}{2}I_2(s)\rightarrow HI(g)$     $\Delta H_1=26kJ/mol$

(2) $HI(g)\rightarrow \frac{1}{2}H_2(g)+\frac{1}{2}I_2(g)$     $\Delta H_2=5.0kJ/mol$

The expression for enthalpy of sublimation of iodine will be,

$\Delta H=\Delta H_1+\Delta H_2$

$\Delta H=(26kJ/mol)+(5.0kJ/mol)$

$\Delta H=31kJ/mol$

Therefore, the enthalpy change for the sublimation of iodine is, 31 kJ/mol