Predict in which of the following,entropy increases or decreases:
$(i)$ $A$ liquid crystallizes into a solid.
$(ii)$ Temperature of a crystalline solid is raised from $0 \, K$ to $115 \, K$.
$(iii)$ $2NaHCO_{3(s)} \to Na_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$
$(iv)$ $H_{2(g)} \to 2H_{(g)}$

(A) $(i)$ After freezing,the molecules attain an ordered state,therefore,entropy decreases.
$(ii)$ At $0 \, K$,the constituent particles are static and entropy is minimum. If the temperature is raised to $115 \, K$,these particles begin to move and oscillate about their equilibrium positions in the lattice,and the system becomes more disordered. Therefore,entropy increases.
$(iii)$ The reactant,$NaHCO_3$,is a solid and has low entropy. Among the products,there is one solid and two gases. Therefore,the products represent a condition of higher entropy.
$(iv)$ Here,one molecule gives two atoms,i.e.,the number of particles increases,leading to a more disordered state. Two moles of $H$ atoms have higher entropy than one mole of $H_2$ molecule.