Question

Calculate the standard molar enthalpy of formation of NO(g) from the following data:

\({{\bf{N}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{ + 2}}{{\bf{O}}_{\bf{2}}} \to {\bf{2N}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{\;\;\;\;\;\Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}{\bf{ = 66}}{\bf{.4kJ}}\)

\({\bf{2NO}}\left( {\bf{g}} \right){\bf{ + }}{{\bf{O}}_{\bf{2}}} \to {\bf{2N}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{\;\;\;\;\Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}{\bf{ = - 114}}{\bf{.1kJ}}\)

Short answer

The change in enthalpy of the given chemical equation will be 90.25 kJ. A positive enthalpy change means that the energy is absorbed during the reaction.

Step-by-step solution

Use Hess’s law

For solving the given problem, we will use the Hess’s law and follow the steps below:

\({{\bf{N}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{ + 2}}{{\bf{O}}_{\bf{2}}} \to {\bf{2N}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{\;\;\;\;\;\Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}{\bf{ = 66}}{\bf{.4kJ}}\)

\({\bf{2NO}}\left( {\bf{g}} \right){\bf{ + }}{{\bf{O}}_{\bf{2}}} \to {\bf{2N}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{\;\;\;\;\Delta H}}_{{\bf{298 }}}^{\bf{^\circ }}{\bf{ = - 114}}{\bf{.1kJ}}\)

Multiply both the equations by 1/2 and replace equation (2) with equation (1).

Change in enthalpy

The change in enthalpy is calculated as shown below.

1/2N₂(g) + 1/2 O₂(g) →NO(g)............(3)

ΔH⁰₂₉₈ = (1/2 x 66.4 + 1/2 x 114.1)kJ

= (33.2+57.05) kJ

= 90.25kJ

Hence, the change in enthalpy of the given chemical equation will be 90.25 kJ.