Question

The standard heats of formation of $N{O_2}(g)$ and ${N_2}{O_4}(g)$ are $33.5$ and $8.4\;kJmo{l^{ - 1}}$ respectively. The heat of dimerization of $N{O_2}$ in kJ is:
A. $41.9$
B. $ - 25.1$
C. $ - 52.2$
D. $ - 58.6$

Answer 1

Hint: To find the heat of dimerization of $N{O_2}$ molecule, the Hess’s law will be applicable which states that regardless of the multiple steps or stages of a reaction, the total enthalpy change for a specific reaction is the sum of all changes that means enthalpy of a reaction is a state function.

Complete answer:
Standard heat of formation: It is defined as the change in enthalpy when one mole of a substance in the standard state is formed from its pure elements. The standard enthalpy of formation for $N{O_2}$ and ${N_2}{O_4}$ can be given as per following reactions:
$\dfrac{1}{2}{N_2}(g) + {O_2}(g) \to N{O_2}(g)\;\;\;\;\;\Delta H_f^o = 33.5\;kJmo{l^{ - 1}}\;\;\;\;\;\;\;\;...(1)$
${N_2}(g) + 2{O_2}(g) \to {N_2}{O_4}(g)\;\;\;\;\Delta H_f^o = 8.4\;kJmo{l^{ - 1}}\;\;\;\;\;\;\;\;\;\;...(2)$
Now, we know that dimerization is an addition reaction in which two molecules of the same compound react together to form a product which is usually referred to as adduct. The dimerization of $N{O_2}$ takes place as follows:
$2N{O_2} \rightleftharpoons {N_2}{O_4}$
On applying Hess’s law, the heat of dimerization of $N{O_2}$ will be equal to $\Delta H_{{f_2}}^o - 2\Delta H_{{f_1}}^o$. On substituting values, the heat of dimerization of $N{O_2}$ will be as follows:
$ \Rightarrow \Delta {H_d} = 8.4 - 2 \times 33.5$
$ \Rightarrow \Delta {H_d} = - 58.6\;kJ$
Hence, the heat of dimerization of $N{O_2}$ is $ - 58.6$.
So, option (D) is the correct answer.

Note:
It is important to note that the sign of the enthalpy of a reaction changes when the process is reversed. Since enthalpy is a state function i.e., it does not depend on the path. Therefore, it does not matter what reactions are used to obtain the enthalpy of the final reaction.