Question

The equilibrium constant for reaction H2 +I2 gives 2HI is 32 at a given temperature the equilibrium constant of I2 and HI are 0.5×10'*-3 and 8×10*-3M the equilibrium constant of H2is

Answer 1

Answer : The equilibrium concentration of $H_2$ is, $4\times 10^{-3}M$

Solution : Given,

Equilibrium constant = 32

Equilibrium concentration of $I_2$ = $0.5\times 10^{-3}M$

Equilibrium concentration of $HI$ = $8\times 10^{-3}M$

The given balanced equilibrium reaction is,

$H_2+I_2\rightleftharpoons 2HI$

The expression for equilibrium constant will be,

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

Now put all the given values in this formula, we get

$32=\frac{(8\times 10^{-3}M)^2}{[H_2]\times (0.5\times 10^{-3}M)}$

$[H_2]=4\times 10^{-3}M$

Therefore, the equilibrium concentration of $H_2$ is, $4\times 10^{-3}M$

Answer 2

Answer:

Answer : The equilibrium concentration of H_2H

2

is, 4\times 10^{-3}M4×10

−3

M

Solution : Given,

Equilibrium constant = 32

Equilibrium concentration of I_2I

2

= 0.5\times 10^{-3}M0.5×10

−3

M

Equilibrium concentration of HIHI = 8\times 10^{-3}M8×10

−3

M

The given balanced equilibrium reaction is,

H_2+I_2\rightleftharpoons 2HIH

2

+I

2

⇌2HI

The expression for equilibrium constant will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}K

c

=

[H

2

][I

2

]

[HI]

2

Now put all the given values in this formula, we get

32=\frac{(8\times 10^{-3}M)^2}{[H_2]\times (0.5\times 10^{-3}M)}32=

[H

2

]×(0.5×10

−3

M)

(8×10

−3

M)

2

[H_2]=4\times 10^{-3}M[H

2

]=4×10

−3

M

Therefore, the equilibrium concentration of H_2H

2

is, 4\times 10^{-3}M4×10^−3M