Equilibrium constant of a reaction, $3 F e (s) + 4 H_{2} O_{(g)} ⇌ F e_{3} O_{4} (s) + 4 H_{2 (g)}$ is same whether, the reaction is carried out in an open vessel or a closed vessel.

Open in App

Solution

The given statement is false. Since in open vessel $H_{2}$ escapes there is no equilibrium attained. Hence we should not talk about equilibrium constant.

Suggest Corrections

Similar questions

Q. For the equilibrium reaction

3 F e (s) + 4 H_{2} O (g) ⇋ F e_{3} O_{4} (s) + 4 H_{2} (g)

, the relationship between

K_{p}

and

K_{c}

is:

Q. In the equilibrium

4 H_{2} O (g) + 3 F e (s) ⇋ F e_{3} O_{4} (s) + 4 H_{2} (g)

, the yield of

H_{2}

can be increased by:

3 F e_{(s)} + 4 H_{2} O_{(g)} ⇌ F e_{3} O_{4 (s)} + 4 H_{2 (g)} +

heat. Which of the following effects disturb the system from equilibrium?

Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a)

(b)

(c)

Q. Does the number of moles of reaction products increase, decrease or remain same when each of the following equiibria is subjected to a decrease in pressure bp increasing the volume?
i)

P C l_{5} ⇌ P C l_{3} (g) + C l_{2} (g)

ii)

C a O (s) + C O_{2} (g) ⇌ C a C O_{3} (s)

iii)

3 F e (s) + 4 H_{2} O (g) ⇌ F e_{3} O_{4} (s) + 4 H_{2} (g)

Join BYJU'S Learning Program

Equilibrium constant of a reaction, 3Fe(s)+4H2O(g)⇌Fe3O4(s)+4H2(g) is same whether, the reaction is carried out in an open vessel or a closed vessel.

The given statement is false. Since in open vessel H2 escapes there is no equilibrium attained. Hence we should not talk about equilibrium constant.

Equilibrium constant of a reaction, $3 F e (s) + 4 H_{2} O_{(g)} ⇌ F e_{3} O_{4} (s) + 4 H_{2 (g)}$ is same whether, the reaction is carried out in an open vessel or a closed vessel.

The given statement is false. Since in open vessel $H_{2}$ escapes there is no equilibrium attained. Hence we should not talk about equilibrium constant.