Change in entropy is negative for

Question

Change in entropy is negative for (A) Bromine (ℓ)→ Bromine (g) (B) C(s) + H2O(g)→ CO(g) + H2(g) (C) N2(g, 10 atm) → N2(g, 1 atm) (D) Fe (1 mol, 400 K)→ Fe (1 mol, 300 K)

Tardigrade Admin · Accepted Answer

Δ S= Sigma n Δ S (Products) - Sigma n Δ S (Reactants) In option (d) change in entropy is negative because at higher temperature degree of randomness is high.

Q. Change in entropy is negative for

Bromine $(ℓ) \to$ Bromine $(g)$

$C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$

$N_{2} (g, 10 a t m) \to N_{2} (g, 1 a t m)$

$F e (1 m o l, 400 K) \to F e (1 m o l, 300 K)$

$Δ S = Σ n Δ S$ (Products) $- Σ n Δ S$ (Reactants) In option (d) change in entropy is negative because at higher temperature degree of randomness is high.

Q. Change in entropy is negative for

Bromine (ℓ)→ Bromine (g)

C(s)+H2​O(g)→CO(g)+H2​(g)

N2​(g,10atm)→N2​(g,1atm)

Fe(1mol,400K)→Fe(1mol,300K)

ΔS=ΣnΔS (Products) −ΣnΔS (Reactants) In option (d) change in entropy is negative because at higher temperature degree of randomness is high.

Bromine $(ℓ) \to$ Bromine $(g)$

$C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$

$N_{2} (g, 10 a t m) \to N_{2} (g, 1 a t m)$

$F e (1 m o l, 400 K) \to F e (1 m o l, 300 K)$

$Δ S = Σ n Δ S$ (Products) $- Σ n Δ S$ (Reactants) In option (d) change in entropy is negative because at higher temperature degree of randomness is high.