Equilibrium concentration of HI, I2 and H2 are 0.7, 0.1 and 0.1 M respectively. The value of equilibrium constant for the reaction, H2+I2leftharpoons 2HI is:

Question

Equilibrium concentration of HI, I2 and H2 are 0.7, 0.1 and 0.1 M respectively. The value of equilibrium constant for the reaction, H2+I2leftharpoons 2HI is: (A) 36 (B) 49 (C) 0.49 (D) 0.36

Tardigrade Admin · Accepted Answer

beginalign beginmatrix H2+ I2leftharpoons 2HI textinitial textconc text. a a 0 textconc text. textat textequil text. (a-x) (a-x) x=0.7 endmatrix beginmatrix =0.1 =0.1 endmatrix endalign ∴ Kc =([HI]2/[H2] [I2]) =((0.7)2/(0.1) (0.1)) =(0.49/0.01) =49

Q. Equilibrium concentration of $H I, I_{2}$ and $H_{2}$ are 0.7, 0.1 and 0.1 M respectively. The value of equilibrium constant for the reaction, $H_{2} + I_{2} ⇌ 2 H I$ is:

36

49

0.49

0.36

$initial conc . conc . at equil . H_{2} + a (a - x) I_{2} ⇌ a (a - x) 2 H I 0 x = 0.7 = 0.1 = 0.1$ $∴ K_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{( 0.7 ) ^{2}}{( 0.1 ) ( 0.1 )}$ $= \frac{0.49}{0.01} = 49$

Q. Equilibrium concentration of HI,I2​ and H2​ are 0.7, 0.1 and 0.1 M respectively. The value of equilibrium constant for the reaction, H2​+I2​⇌2HI is:

36

49

0.49

0.36

​initialconc.conc.atequil.​H2​+a(a−x)​I2​⇌a(a−x)​2HI0x=0.7​=0.1​=0.1​​ ∴Kc​=[H2​][I2​][HI]2​=(0.1)(0.1)(0.7)2​ =0.010.49​=49

Q. Equilibrium concentration of $H I, I_{2}$ and $H_{2}$ are 0.7, 0.1 and 0.1 M respectively. The value of equilibrium constant for the reaction, $H_{2} + I_{2} ⇌ 2 H I$ is:

$initial conc . conc . at equil . H_{2} + a (a - x) I_{2} ⇌ a (a - x) 2 H I 0 x = 0.7 = 0.1 = 0.1$ $∴ K_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{( 0.7 ) ^{2}}{( 0.1 ) ( 0.1 )}$ $= \frac{0.49}{0.01} = 49$