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) H = -92.kJ

In the second step, ammonia and oxygen react to form nitric acid and water:

NH3(g) + 2O2(g) ----> HNO3(g) + H20 (g) H=-330.kJ

Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen. Round your answer to the nearest kJ.

Answer 1

Answer: Hence, the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen is -376 kJ

Step-by-step 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)+3H_2(g)\rightarrow 2NH_3(g)` (1)

`\Delta H_1=-92kJ`

`NH_3(g)+2O_2(g)\rightarrow HNO_3(g)+H_2O(g)` (2)

`\Delta H_2=-330kJ`

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

`(1)/(2)N_2(g)+(3)/(2)H_2(g)+2O_2(g)\rightarrow HNO_3(g)+H_2O`
`\Delta H_3=?` (3)

According to the Hess’s law, if we multiply the reaction 1 by
`(1)/(2)` and add to 2.

`\Delta H_3=(1)/(2)* (-92kJ)+(-330kJ)`

`\Delta H_3=-376kJ`

Hence, the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen is -376 kJ