Question

Enter your answer in the provided box. Oxidation of gaseous ClF by F2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH o rxn for the production of ClF3: (1) 2 ClF(g) + O2(g) → Cl2O(g) + OF2(g) ΔHo = 167.5 kJ (2) 2 F2(g) + O2(g) → 2 OF2(g) ΔHo = −43.5 kJ (3) 2 ClF3(l) + 2 O2(g) → Cl2O(g) + 3 OF2(g) ΔHo = 394.1 kJ

Answer 1

Answer : The value of $\Delta H_{rxn}$  for the reaction is, -135.2 kJ

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 formation of $ClF_3$ will be,

$ClF(g)+F_2(g)\rightarrow ClF_3(l)$    $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $2ClF(g)+O_2(g)\rightarrow Cl_2O(g)+OF_2(g)$     $\Delta H_1=167.5kJ$

(2) $2F_2(g)+O_2(g)\rightarrow 2OF_2(g)$    $\Delta H_2=-43.5kJ$

(3) $2ClF_3(l)+2O_2(g)\rightarrow Cl_2O(g)+3OF_2(g)$    $\Delta H_3=394.4kJ$

We are dividing the reaction 1, 2 and 3 and reversing reaction 3 and then adding all the equations, we get :

(1) $ClF(g)+\frac{1}{2}O_2(g)\rightarrow \frac{1}{2}Cl_2O(g)+\frac{1}{2}OF_2(g)$     $\Delta H_1=\frac{167.5kJ}{2}=83.75kJ$

(2) $F_2(g)+\frac{1}{2}O_2(g)\rightarrow OF_2(g)$    $\Delta H_2=\frac{-43.5kJ}{2}=-21.75kJ$

(3) $\frac{1}{2}Cl_2O(g)+\frac{3}{2}OF_2(g)\rightarrow ClF_3(l)+O_2(g)$    $\Delta H_3=\frac{-394.4kJ}{2}=-197.2kJ$

The expression for enthalpy of formation of $ClF_3$ will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(83.75kJ)+(-21.75kJ)+(-197.2kJ)$

$\Delta H_{rxn}=-135.2kJ$

Therefore, the value of $\Delta H_{rxn}$  for the reaction is, -135.2 kJ