Question

Which one of the following has $△S^o$ greater than zero ?

• A. $CaO\left(s\right)+C{O}_2\left(g\right)\rightleftharpoons CaC{O}_3\left(s\right)$
• B. $NaCl\left(aq\right)\rightleftharpoons NaCl\left(s\right)$
• C. $NaN{O}_3\left(s\right)\rightleftharpoons N{a}^{+}\left(aq\right)+N{O}_3^{-}\left(aq\right)$
• D. $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

Answer 1

The correct option is C $NaN{O}_3\left(s\right)\rightleftharpoons N{a}^{+}\left(aq\right)+N{O}_3^{-}\left(aq\right)$

$NaN{O}_3$ is a solid, which is converted to liquid ions. As we know that, in transition from solid state to liquid state, entropy increases. Hence the value of $\Delta S$ will be greater than zero for the reaction- $NaN{O}_3\left(s\right)\rightleftharpoons N{a}^{+}(aq.)+N{O}_3^{-}(aq.)$.
In option a , $CaO$ combines with $C{O}_2$ , So from 2 molecules of reactant 1 molecules of product is forming i.e. entropy decreases.
In option b , liquid to solid $NaCl$ is forming , So here also entropy decreases.
In option d , total 4 moles of reactant is forming 2 moles of products. Therefore , from reactant to product entropy decreases.