Question

If $\Delta H$ value for the manufacture of $N{H_3}$ is $\Delta H = - 91.8kJ$. The correct thermochemical equation for the dissociation of $N{H_3}$ is
A.${N_2} + 3{H_2} - 91.8 \to 2N{H_3}$
B.$2N{H_3} \to {N_2} + 3{H_2}$ $\Delta H = + 91.8kJ$
C.\[2N{H_3} \to {N_2} + 3{H_2}\] \[\Delta H = - 91.8kJ\]
D.$\dfrac{1}{2}{N_2} + \dfrac{3}{2}{H_2} \to 2N{H_3}$ $\Delta H = + 91.8kJ$

Answer 1

Hint: We can say that the thermochemical equation is a balanced stoichiometric chemical equation that contains $\Delta H$ which is nothing but change in enthalpy. The value of change in enthalpy could be either positive (or) negative. The unit of change in enthalpy is $kJ$.

Complete step by step solution:
Now we can discuss about the concept of endothermic reaction as,
Endothermic reaction: Those reactions that absorb heat energy from the surroundings are called endothermic reactions. The breakage of chemical bonds absorbs heat energy. Endothermic reactions always have a positive enthalpy because they need energy for the reaction to proceed.
Example: In the process of photosynthesis, carbon dioxide and water are converted to free oxygen and glucose by absorbing heat energy.
$6C{O_2} + 6{H_2}O + 2519kJ \to {C_6}{H_{12}}{O_6} + 6{O_2}$
Now we discuss about the concept of exothermic reaction as,
Exothermic reaction: Those reactions that release heat energy to the surroundings are called exothermic reactions. The formation of chemical bonds, releases heat energy. Exothermic reactions always have negative enthalpy because they release energy upon the completion of the reaction.
Example: In the combination reaction of solid carbon and gaseous oxygen, formation of gaseous carbon dioxide takes place with the evolution of heat energy.
$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right) + 393kJ$
We know that the formation of ammonia has an enthalpy change of $ - 91.8kJ$. The negative value of enthalpy indicates that the formation of ammonia is an exothermic reaction. So, we can write the thermochemical equation for the formation of ammonia as,
${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$ $\Delta H = - 91.8kJ$
We can say that the dissociation of ammonia is a reverse reaction of formation ammonia. The reaction would have an enthalpy change of $ + 91.8kJ$. The positive value of change in enthalpy indicates that the dissociation of ammonia is an endothermic reaction. So, we can write the thermochemical equation for the dissociation of ammonia as,
$2N{H_3}\left( g \right) \to {N_2}\left( g \right) + 3{H_2}\left( g \right)$ $\Delta H = + 91.8kJ$
From the thermochemical equation of dissociation of ammonia, we can see that Option (B) is correct.

Note:
We have to know that energy changes that accompany chemical reactions are indicated by the thermochemical equations. We can say a reaction is endothermic, if the energy is in the reactant side, and if the energy is in the product, then the reaction is exothermic. A thermochemical equation should also contain the physical states of the products and reactants other than changes in energy.