Question

Nitric acid can be manufactured in a multi-step process, during which nitric oxide is oxidized to create nitrogen dioxide. 2NO (g) + O2 (g) → 2NO2 (g)
Calculate the standard reaction enthalpy for the above reaction using the following thermodynamic data.
N2 (g) + O2 (g) → 2NO (g) ∆H˚1 = 180.5 kJ
N2 (g) + 2O2 (g) → 2NO2 (g) ∆H˚2 = 66.4 kJ
-252.4 kJ/mol rxn
-114.1 kJ/mol rxn
-100.3 kJ/mol rxn
-246.9 kJ/mol rxn

Answer 1

Answer: The standard reaction enthalpy for the above reaction is -114.1 kJ/mol

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 chemical reactions are,

$N_2(g)+O_2(g)\rightarrow 2NO(g)$ (1) $\Delta H_1=180.5kJ$

$N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$ (2) $\Delta H_2=66.4kJ$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$ (3) $\Delta H_3=?$

Reversing reaction (1)

$2NO(g)\rightarrow N_2(g)+O_2(g)$ (1') $\Delta H_1'=-180.5kJ$

Adding reaction 1' and 2

So, the value $\Delta H_3$ for the reaction will be:

$\Delta H_3=\Delta H_1'+\Delta H_2$

$\Delta H_3=-180.5+66.4=-114.1kJ$

Hence, the standard reaction enthalpy for the above reaction is -114.1 kJ/mol