For the reaction 2 Cl ( g ) rightarrow Cl _{2 | vedclass

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Question

(A) In the reaction $2 Cl ( g ) \rightarrow Cl _{2}( g )$,two gaseous atoms combine to form one molecule of chlorine gas.$1$. Enthalpy change $(\Delta

Vedclass · Accepted Answer

$\Delta_{ r } H < 0$ and $\Delta_{ r } S < 0$

Vedclass · Answer

(A) In the reaction $2 Cl ( g ) \rightarrow Cl _{2}( g )$,two gaseous atoms combine to form one molecule of chlorine gas.$1$. Enthalpy change $(\Delta_{ r } H)$: Bond formation is an exothermic process,therefore energy is released,making $\Delta_{ r } H $2$. Entropy change $(\Delta_{ r } S)$: The number of moles of gas decreases from $2$ to $1$ $(2 Cl ( g ) \rightarrow 1 Cl _{2}( g ))$. Since the randomness or disorder of the system decreases,$\Delta_{ r } S Thus,both $\Delta_{ r } H < 0$ and $\Delta_{ r } S < 0$.

For the reaction $2 Cl ( g ) \rightarrow Cl _{2}( g )$,the correct option is

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