Question

Which of the following most likely to proceed with a decrease in entropy?

• A. ${Ag}^{+}\left(aq\right)+{Cl}^{-}\to AgCl\left(aq\right)$
• B. $C_5{H}_{12}\left(s\right)+8O_2\left(g\right)\to 5{CO}_2\left(g\right)+6H_{2}O\left(l\right)$
• C. ${NH}_3\left(s\right)+H_{2}O\left(s\right)\leftrightharpoons {NH}_4^{+}\left(s\right)+{OH}^{-}\left(s\right)$
• D. $H{C}_2{H}_3{O}_2\left(l\right)+{OH}^{-}\left(aq\right)\to H_{2}O\left(O\right)+C_2{H}_3{O}_2^{-}\left(aq\right)$
• E. $Ca{CO}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)$

Answer 1

The correct option is A ${Ag}^{+}\left(aq\right)+{Cl}^{-}\to AgCl\left(aq\right)$

Option A is the correct answer.

In this equation, there are $two$ moles of ions on the left-hand side of the equation ($A{g}^{+}$ and $C{l}^{-}$). When they react they form only $1$ mole of $AgCl$, which is a white precipitate.

$A{g}^{+}$ $+$ $C{l}^{-}$ $\to$$AgCl$

In all other equations, the number of moles of products are either same or increases than the number of moles of reactants.

Entropy means randomness. Since, in option A, the number of moles of products (which is one) is less than the number of moles of reactants (which is two). Hence, the randomness or entropy of the system in option A will decrease the formation of the product as the number of moles decreases.