Question

Explain with reason, sign convention of $\Delta S$ in the following reaction.

$N_{2\left(g\right)}+3H_{2\left(g\right)}⟶2N{H}_{3\left(g\right)}$

Answer 1

$N_{2\left(g\right)}+3H_{2\left(g\right)}\to 2N{H}_{3\left(g\right)}$

(Total moles of gaseous reactants) > (Total moles of gaseous products)

$\therefore$ (Disordered state of gaseous reactants) > (Disordered state of gaseous products)

In the reaction, 4 mole of gaseous reactants form 2 moles of gaseous products, i.e., $\Delta$ n > 0.

Therefore, disorder decreases and hence entropy decreases ($\Delta$ S < 0).

So, sign of $\Delta S$ is $-ve$.