Question

Calculate ΔHfor the process:

\({{\bf{N}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

from the following information:

\({{\bf{N}}_{\bf{2}}}{\bf{(g) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2NO(g)}};{\bf{ \Delta H = 180}}{\bf{.5 kJ}}\)

\({\bf{NO(g) + 1/2 }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}};{\bf{ \Delta H = - 57}}{\bf{.06 kJ}}\)

Short answer

ΔH = 66.4 kJ.

Step-by-step solution

Enthalpy change

ΔHis termed enthalpy change and it refers to the amount of heat that gets released or absorbed during a chemical reaction. Its unit is kilojoules (kJ).

Calculation of enthalpy change

The reaction for which we have to find the ΔHis the sum of the following reactions given below with theirΔH.

To balance the reaction, we have to multiply reaction (b) by 2. Hence,

\(\begin{array}{*{20}{l}}{{\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + }}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2NO(g); \Delta H = 180}}{\rm{.5 kJ\;}}.......{\rm{ }}\left( {\rm{a}} \right)}\\{\rm{\;}}\\{{\rm{2NO(g) + \;}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2N}}{{\rm{O}}_{\rm{2}}}{\rm{(g); \Delta H = - 57}}{\rm{.06 kJ\;}} \times {\rm{2 }}......\left( {\rm{b}} \right)}\end{array}\)

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\({{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2N}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}\)

Thus, the enthalpy change for the reaction

\({{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2N}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}\)is 66.38 kJ, that is, 66.4 kJ.