Starting from -a- moles of H2 and -b- moles of I2 an equilibrium H2 - I2 - 2HI is established with 2x moles of HI- The equilibrium constant Kc is -

The correct option is B 4x ^ 2 /( a x ) ( b x ) H _ 2 + I _ 2 ⇌ 2HI at t _ 1 =0 a #160;b #160; #160;at equilibrium a x #160;b x ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

Starting from...

Question

Starting from 'a' moles of $H_{2}$ and $^{'} b^{'}$ moles of $I_{2}$ an equilibrium $H_{2} + I_{2} \Leftrightarrow 2 H I$ is established with 2x moles of HI. The equilibrium constant $K_{c}$ is ?

\frac{4 x^{2}}{a b}

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\frac{{4 x}^{2}}{(a - x) (b - x)}

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\frac{2 x}{(a - x) (b - x)}

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\frac{{4 x}^{2}}{(a - 2 x) (b - 2 x)}

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Solution

The correct option is B $\frac{{4 x}^{2}}{(a - x) (b - x)}$
$H_{2} + I_{2} ⇌ 2 H I$
at $t_{1} = 0$ a b -
at equilibrium a-x b-x 2x
$K_{c} = \frac{[2 x]^{2}}{[a - x] [b - x]} = \frac{{4 x}^{2}}{[a - x] [b - x]}$

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Similar questions

Q. Evaluate

K_{p}

for the reaction

: H_{2} + I_{2} ⇌ 2 H I

. If

2

moles each of

H_{2}

and

I_{2}

are taken initially. The equilibrium moles of

H I

2

Q. At certain temperature, a

10

lit vessel contains

0.4

mole

H_{2}

0.4

mole

I_{2}

and

0.1

mole

H I

at equilibrium. Then

K_{p}

for

H_{2} + I_{2} ⇌ 2 H I

is:

Q. In the reaction,

H_{2} + I_{2} ⇌ 2 H L

. In a

2

litre flask

0.4

moles of each

H_{2} & I_{2}

are taken. AT equilibrium

0.5

moles of HI are formed. What will be the value of equilibrium constant,

K_{c}

Q. On a given condition , the equilibrium concentration of

H I

H_{2}

and

I_{2}

are

0.80, 0.10 a n d 0.10 m o l e / l i t r e

the equilibrium constant for the reaction

H_{2}

I_{2}

⇌ 2 H I

will be:

Q. One mole of

H_{2}

2

moles of

I_{2}

and

3

moles of

H I

are injected in a one litre flask. What will be the concentration of

H_{2}, I_{2}

and

H I

at equilibrium when

K_{c}

45.9

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Starting from 'a' moles of H2 and ′b′ moles of I2 an equilibrium H2+I2⇔2HI is established with 2x moles of HI. The equilibrium constant Kc is ?

The correct option is B 4x2(a−x)(b−x)H2+I2⇌2HIat t1=0 a b - at equilibrium a-x b-x 2x Kc=[2x]2[a−x][b−x]=4x2[a−x][b−x]

Starting from 'a' moles of $H_{2}$ and $^{'} b^{'}$ moles of $I_{2}$ an equilibrium $H_{2} + I_{2} \Leftrightarrow 2 H I$ is established with 2x moles of HI. The equilibrium constant $K_{c}$ is ?

The correct option is B $\frac{{4 x}^{2}}{(a - x) (b - x)}$
$H_{2} + I_{2} ⇌ 2 H I$
at $t_{1} = 0$ a b -
at equilibrium a-x b-x 2x
$K_{c} = \frac{[2 x]^{2}}{[a - x] [b - x]} = \frac{{4 x}^{2}}{[a - x] [b - x]}$