Question

Value of Δ H ∘ rxn for the equation NH 3 ( g ) + 2 O 2 ( g ) ⟶ HNO3 ( g ) + H 2 O ( g )

Answer 1

Answer: - 894.6 kJ/mol.

Explanation:

Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

∆H is the change in the sum of the internal energy of a system.

We are to find the Value of ΔH°(rxn) for the equation:

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).

===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

Δ H^∘ (rxn)= - 894.6 kJ/mol.