Calculate δh (in kj) for the process hg2br2(s) → 2 hg(l) + br2(l) from the following information.
hg(l) + br2(l) → hgbr2(s) δh⁰298 = −170.7 kj
hg(l) + hgbr2(s) → hg2br2(s) δh⁰298 = −36.2 kj

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,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

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

(1)    

(2)    

The expression for final enthalpy is,

Therefore, the enthalpy change of reaction is 206.9 kJ

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.