Change in entropy is negative for:

B r o m i n e (I) \to B r o m i n e (g)

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

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

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F e (a t 400 K) \to F e (a t 300 K)

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Solution

The correct option is B $B r o m i n e (I) \to B r o m i n e (g)$
When $B r (I)$ goes to to $B r (g)$ then stability of molecules increase. This leads us to negative entropy.

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Similar questions

Calculate the equilibrium constant ( $K_{p}$ ) for the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , at $1300 K$ from the following data:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$

$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$

Q. For the reaction,

C (s) + H_{2} O (g) ⇌ H_{2} (g) + C O (g)

, by increasing pressure following equilibrium :

Q. For which of the following processes is

Δ S

negative?
a.

H_{2} (g) ⟶ 2 H (g)

N_{2} (g, 1 a t m) ⟶ N_{2} (g, 5 a t m)

2 S O_{3} (g) ⟶ 2 S O_{2} (g) + O_{2} (g)

C (d i a m o n d) ⟶ C (g r a p h i t e)

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)

The above equilibrium when subjected to pressure:

C_{(s)} + H_{2} O_{(g)} \to C O (g) + H_{2} (g)

The pressure on the system is increased in the above reaction then:

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Change in entropy is negative for:

The correct option is B Bromine(I)→Bromine(g)When Br(I) goes to to Br(g) then stability of molecules increase. This leads us to negative entropy.

The correct option is B $B r o m i n e (I) \to B r o m i n e (g)$
When $B r (I)$ goes to to $B r (g)$ then stability of molecules increase. This leads us to negative entropy.