Question

Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2(g) + 3H2(g) 2NH3(g) AH=-92. kJ In the second step, ammonia and oxygen react to form nitric acid and water: NH3(9) + 2O2(g) → HNO3(9) + H2O(g) AH=-330. kJ Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest kJ. ПkJ 1 x Ś ?

Answer 1

Explanation:

According to the Hess's law of constant heat summation, the total enthalpy change for a reaction is the sum of all changes regardless of multiple stages or steps in a reaction.

As the give first reaction equation is as follows.

       $N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)$,   $\Delta H = -92 kJ$

Dividing this equation by 2 so, the value of enthalpy will also get half. Then the equation will be as follows.

        $\frac{1}{2}N_{2}(g) + \frac{3}{2}H_{2}(g) \rightarrow \frac{2}{2}NH_{3}(g)$,   $\Delta H = \frac{-92}{2} kJ$  

     $\frac{1}{2}N_{2}(g) + \frac{3}{2}H_{2}(g) \rightarrow NH_{3}(g)$,   $\Delta H = -46 kJ$  ............ (1)

The second equation will be as follows.

     $NH_{3}(g) + 2O_2(g) \rightarrow HNO_{3}(g) + H_{2}O(g)$   $\Delta H$ = -330 kJ ............ (2)

On adding both equation (1) and (2), the net reaction equation and enthalpy for 1 mole of nitric acid will be as follows.

      $\frac{1}{2}N_{2}(g) + \frac{3}{2}H_{2}(g) + 2O_{2}(g) \rightarrow HNO_{3}(g) + H_{2}O(g)$  

    [tex]\Delta H = [-330 + (-46)] kJ    

              = -376 kJ

Thus, we can conclude that the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions is -376 kJ.