Question

Calculate ΔH (in kJ) for the process Hg2Br2(s) → 2 Hg(l) + Br2(l) from the following information.

Answer 1

Answer : The enthalpy change of reaction is 206.9 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 given final reaction is,

$Hg_2Br_2(s)\rightarrow 2Hg(l)+Br_2(l)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $Hg(l)+Br_2(l)\rightarrow HgBr_2(s)$    $\Delta H^o_1=-170.7kJ$

(2) $Hg(l)+HgBr_2(s)\rightarrow Hg_2Br_2(s)$    $\Delta H^o_2=-36.2kJ$

First we will reverse the reaction 1 and 2 then adding both the equation, we get :

(1) $HgBr_2(s)\rightarrow Hg(l)+Br_2(l)$    $\Delta H^o_1=+170.7kJ$

(2) $Hg_2Br_2(s)\rightarrow Hg(l)+HgBr_2(s)$    $\Delta H^o_2=+36.2kJ$

The expression for final enthalpy is,

$\Delta H=\Delta H^o_1+\Delta H^o_2$

$\Delta H=(+170.7kJ)+(+36.2kJ)$

$\Delta H=206.9kJ$

Therefore, the enthalpy change of reaction is 206.9 kJ

Answer 2

Answer: $\Delta H^o_3$ is 206.3 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.

$Hg(l)+Br_2(l)\rightarrow HgBr_2(s)$    $\Delta H^o_1=-170.7kJ$   ......(1)

$Hg(l)+HgBr2(s)\rightarrow Hg_2Br_2$ $\Delta H^o_2=-36.2kJ$  .......(2)

The final reaction is:  

$HgBr_2(s)\rightarrow 2Hg(l)+Br_2(l)$  $\Delta H^o_3=?$   ......(3)

By reversing both the equations and then adding them, we get:

$\Delta H^o_3=-\Delta H^o_1+(-\Delta H^o_2)=-(-170.7)+[-(-36.2kJ)]=206.3kJ$

Hence, $\Delta H^o_3$ is 206.3 kJ.