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For the reaction \(2 C l(g) \rightarrow C l_{2}(g)\), the correct option is:
A \(\Delta_{r} H>0\) and \(\Delta_{r} S>0\)
в \(\Delta_{r} H>0\) and \(\Delta_{r} S<0\)
c \(\Delta_{r} H<0\) and \(\Delta_{r} S>0\)
(D) \(\Delta_{r} H<0\) and \(\Delta_{r} S<0\)

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Solution:
$$
2 \mathrm{Cl}(g) \longrightarrow \mathrm{Cl}_{2}(g)
$$
As the process involves bond formation \(\Delta H=-v e\) and as the number of gaseous particles are decreasing in the reaction \(\Delta S\) is also \(-v e\) i.e., \(\Delta_{r} H<0\) and \(\Delta_{r} S<0\)
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