Question

N2(g) + O2(g) → NO(g) ΔHrxn = +90.3 kJ NO(g) + Cl2(g) → NOCl(g) ΔHrxn = –38.6 kJ What is the value of ΔHrxn for the decomposition of NOCl? 2NOCl(g) → N2(g) + O2(g) + Cl2(g) ΔHrxn = ?

Answer 1

Answer : The value of $\Delta H_{rxn}$ is, -103.4 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 main reaction is,

$2NOCl(g)\rightarrow N_2(g)+O_2(g)+Cl_2(g)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightarrow NO(g)$     $\Delta H_1=+90.3kJ$

(2) $NO(g)+\frac{1}{2}Cl_2(g)\rightarrow NOCl(g)$    $\Delta H_2=-38.6kJ$

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

(1) $2NO(g)\rightarrow N_2(g)+O_2(g)$     $\Delta H_1=2\times (-90.3kJ)=-180.6kJ$

(2) $2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)$    $\Delta H_2=2\times (+38.6kJ)=+77.2kJ$

The expression for $\Delta H_{rxn}$ will be,

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

$\Delta H=(-180.6)+(+77.2)$

$\Delta H=-103.4kJ$

Therefore, the value of $\Delta H_{rxn}$ is, -103.4 kJ

Answer 2

Answer:

51.7kJ

Explanation:

using Hess law