Question

Determine the heat of reaction for the process TiO2(s) + 4HCl(g) TiCl4(l) + 2H2(g) + O2(g) using the information given below: Ti(s) + O2(g) TiO2(s) H° = −939.7 kJ 2HCl(g) H2(g) + Cl2(g) H° = −184.6 kJ Ti(s) + 2Cl2(g) TiCl4(l) H° = −804.2 kJ

Answer 1

Answer : The heat of reaction for the process is, 1374.7 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 main chemical reaction is,

$TiO_2(s)+4HCl(g)\rightarrow TiCl_4(l)+2H_2(g)+O_2(g)$    $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $Ti(s)+O_2(g)\rightarrow TiO_2(s)$     $\Delta H_1=-939.7kJ$

(2) $2HCl(g)\rightarrow H_2(g)+Cl_2(g)$    $\Delta H_2=-184.6kJ$

(3) $Ti(s)+2Cl_2(g)\rightarrow TiCl_4(l)$    $\Delta H_3=-804.2kJ$

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

(1) $TiO_2(s)\rightarrow Ti(s)+O_2(g)$     $\Delta H_1=939.7kJ$

(2) $4HCl(g)\rightarrow 2H_2(g)+2Cl_2(g)$    $\Delta H_2=2\times (-184.6kJ)=-369.2kJ$

(3) $TiCl_4(l)\rightarrow Ti(s)+2Cl_2(g)$    $\Delta H_3=804.2kJ$

The expression for heat of reaction for the process is:

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

$\Delta H_{rxn}=(939.7kJ)+(-369.2kJ)+(804.2kJ)$

$\Delta H_{rxn}=1374.7kJ$

Therefore, the heat of reaction for the process is, 1374.7 kJ