 # For the reaction, ${\rm{2Cl(g)}} \to {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}$, the signs of ${\rm{\Delta H}}$ and ${\rm{\Delta S}}$ respectively, are: A. $+ , -$B. $+ , +$C. $- , -$D. $- , +$ Verified
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Hint: Enthalpy change (${\rm{\Delta H}}$) is defined as the total heat content of the system at constant pressure. Entropy is the extent of disorder of randomness in a system. Entropy change (${\rm{\Delta S}}$) of a substance measures the disorder or randomness in a system.

Complete Step by Step Solution:
The given reaction is as: ${\rm{2Cl(g)}} \to {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}$
In the above reaction, the formation of one chlorine molecule can be seen from two chlorine atoms.

In exothermic reactions, heat energy is evolved (released). This lowers the enthalpy. Consequently, the enthalpy of the products becomes less as compared to the enthalpy of reactants. Therefore,
${\rm{\Delta H}} = {{\rm{H}}_{\rm{p}}} - {{\rm{H}}_{\rm{r}}} = \,{\rm{negative}}\,{\rm{(}} - {\rm{)}}$

In endothermic reactions, heat is absorbed. Heat absorbed raises the enthalpy of the products. Consequently, the enthalpy of products becomes greater than that of reactants. Hence, ${\rm{\Delta H}} = {{\rm{H}}_{\rm{p}}} - {{\rm{H}}_{\rm{r}}} = \,{\rm{positive}}\,{\rm{( + )}}$

Since bond formation is taking place in the reaction, this means that the energy is released (exothermic). Hence, change in enthalpy (${\rm{\Delta H}}$) will be negative ${\rm{(}} - )$.
Entropy has a direct relation with the number of moles of gas. If there is an increase in the number of moles on the product side, then the entropy would be higher (positive). But if there is a decrease in the number of moles on the product side, then the entropy would be lower (negative).

Since two moles of chlorine atom (reactant) forms one mole of chlorine molecule (product), therefore the sign of entropy change (${\rm{\Delta S}}$) will be negative ${\rm{(}} - )$. This is because the number of moles on the product side is less than the number of moles on the reactant side.

Hence, both enthalpy change (${\rm{\Delta H}}$) and entropy change (${\rm{\Delta S}}$) are found to be negative ${\rm{(}} - )$.
Therefore, option C is correct.

Note: Absolute value of enthalpy of a system cannot be determined just like the internal energy. The enthalpy of a system is a state function, therefore, the magnitude of enthalpy change (${\rm{\Delta H}}$) depends only on the enthalpies of the initial and final states. Thus, we can write ${\rm{\Delta H}} = {{\rm{H}}_{{\rm{final}}}} - {{\rm{H}}_{{\rm{initial}}}}$. Entropy is also a state function like enthalpy and internal energy. So, it depends upon the final and initial states of a system. Thus, entropy change can be written as; ${\rm{\Delta S}} = {{\rm{S}}_{{\rm{final state}}}} - {{\rm{S}}_{{\rm{initial state}}}}$ when a system undergoes a change from initial state to final state.