Question

The crystalline hydrate cd(no3)2 ⋅ 4h2o(s) loses water when placed in a large, closed, dry vessel at room temperature: cd(no3)2⋅4h2o(s)→ cd(no3)2(s) 4 h2o(g) this process is spontaneous and δh∘ is positive at room temperature.

Answer 1

The given dehydration equation is,

$Cd(NO_{3})_{2}. 4H_{2}O (s) ---> Cd(NO_{3})_{2}(s) + 4 H_{2}O(g)$

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.

This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.

Answer 2

The given dehydration equation is,

$Cd(NO_{3})_{2}. 4H_{2}O (s) ---> Cd(NO_{3})_{2}(s) + 4 H_{2}O(g)$

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.

This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.