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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

Expert verified

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

01

Use Hess’s law

For solving the given problem, we will use the Hess鈥檚 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).

02

Change in enthalpy

The change in enthalpy is calculated as shown below.

1/2N2(g) + 1/2 O2(g) 鈫扤O(g)............(3)

螖贬0298 = (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.

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