The Kc for H2(g)+I2(g)leftharpoons 2HI(g) is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be:

Question

The Kc for H2(g)+I2(g)leftharpoons 2HI(g) is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be: (A) 28 (B) 64 (C) 32 (D) 16

Tardigrade Admin · Accepted Answer

beginmatrix H2(g)+ I2(g)leftharpoons 2HI(g) textat textequilibrium (x-a) (x-a) (2a) endmatrix Kc=([HI]2/[H2] [I2])=(( (2a/V) )2/((x-a))V× (x-a)V =(4a2/(x-a)2) Thus, /Kc) is independent volume for this reaction, hence, Kc will remain equal to 64 (i.e., equal to initial Kc ).

Q. The Kc​ for H2​(g)+I2​(g)⇌2HI(g) is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be:

28

64

32

16

atequilibrium​H2​(g)+(x−a)​I2​(g)⇌(x−a)​2HI(g)(2a)​ Kc​=[H2​][I2​][HI]2​=V(x−a)​×Vx−a​(V2a​)2​ =(x−a)24a2​ Thus, Kc​ is independent volume for this reaction, hence, Kc​ will remain equal to 64 (i.e., equal to initial Kc​ ).

Q. The $K_{c}$ for $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be: