In which of the following processes, entropy increases? (I) A liquid crystallizes into a solid (II) Temperature of a crystalline solid is raised from 0 K to 115 K (III) 2NaHCO3(s) → Na2CO3(s)+CO2(g)+H2O(g) (IV) H2(g) → 2H(g)

Question

In which of the following processes, entropy increases? (I) A liquid crystallizes into a solid (II) Temperature of a crystalline solid is raised from 0 K to 115 K (III) 2NaHCO3(s) → Na2CO3(s)+CO2(g)+H2O(g) (IV) H2(g) → 2H(g) (A) Only (I) (B) (I) and (II) (C) (II), (III) and (IV) (D) (I) and (IV)

Tardigrade Admin · Accepted Answer

(I) After freezing, the molecules attain an ordered state and therefore, entropy decreases
(II) At

0 K

, the constituent particles are static and entropy is minimum. If temperature is raised to

115 K

, these begin to move and oscillate about their equilibrium positions in the lattice and system becomes more disordered, therefore entropy increases
(III) In the reaction,

2 N a H C O_{3 (s)} \to N a_{2} C O_{3 (s)} + C O_{2 (g)} + H_{2} O_{(g)}

the number of gaseous components are increasing hence the entropy of system will increase
(IV)

H_{2 (g)} \to 2 H_{(g)}

In this decomposition, number of particles increase therefore, the randomness and entropy will also increase

Q. In which of the following processes, entropy increases? (I) A liquid crystallizes into a solid (II) Temperature of a crystalline solid is raised from 0K to 115K (III) 2NaHCO3(s)​→Na2​CO3(s)​+CO2(g)​+H2​O(g)​ (IV) H2(g)​→2H(g)​

Only (I)

(I) and (II)

(II), (III) and (IV)

(I) and (IV)

Q. In which of the following processes, entropy increases?
(I) A liquid crystallizes into a solid
(II) Temperature of a crystalline solid is raised from $0 K$ to $115 K$
(III) $2 N a H C O_{3 (s)} \to N a_{2} C O_{3 (s)} + C O_{2 (g)} + H_{2} O_{(g)}$
(IV) $H_{2 (g)} \to 2 H_{(g)}$