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Characteristics of Equilibrium Constant

In an experim...

Question

In an experiment the equilibrium constant for the reaction A+B\leftrightharpoons C+D is K when the initial concentration of A+B each is 0.1 mol $L^{- 1}$ Under the similar conditions in an another experiement if the initial concentration of A and B are taken 2 and 3 mol $L^{- 1}$ respectively then the value of equilibrium constant will be:-

\frac{K}{2}

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K^{2}

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\frac{1}{K}

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Solution

The correct option is A $\frac{K}{2}$
$A + B ⟶ C + D$
Initially, $0.1$ $0.1$
At eqm $0.1 - x$ $0.1 - x$ $x$ $x$
$∴$ Equilibrium constant $(k) = \frac{[C] [D]}{[A] [B]} = \frac{x \times x}{(0.1 - x) (0.1 - x)} = \frac{x^{2}}{(0.1 - x)^{2}}$
Again for $2 n d$ experiment,
Equilibrium constant= $\frac{[C] [D]}{[A] [B]} = \frac{x \times x}{(2 - x) \times (3 - x)} = k / 2$

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Similar questions

Q. In an experiment the equilibrium constant for the reaction

A + B ⇌ C + D

K

when the initially concentration of

A

and

B

each is

0.1 m o l L^{- 1}

A

and

B

are take

2

and

3 m o l L^{- 1}

respectively then the value of equilibrium constant will be:

33. In an experiment the equilibriurm constant for the
reaction A +B=C+ D is K, when the initial
moles of A and B each is 1.0 moles. Under the
similar conditions in an another experiment if the
initial moles of A and B are taken 2 and 3 moles
then the value of equilibrium constant will be:-

Q. The equilibrium constant at 298 K for a reaction

A + B ⇌ C + D

is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of

D

(in mol L

^{- 1}

) will be:

Q. The equilibrium constants at

298 K

for a reaction

A + B ⇌ C + D

is 100. If the initial concentration of all the four species were

1 M

each then equilibrium concentration of

D ({in mol L}^{- 1})

will be:

The equilibrium constant at $298 K$ for a reaction
$A + B ⇌ C + D$ is $100$ . If the initial concentration of all the four species were $1 M$ each, then equilibrium concentration of $D (in mol/l)$ will be:

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The correct option is A K2 A+B⟶C+DInitially, 0.1 0.1 At eqm 0.1−x 0.1−x x x∴ Equilibrium constant (k)=[C][D][A][B]=x×x(0.1−x)(0.1−x)=x2(0.1−x)2Again for 2nd experiment,Equilibrium constant= [C][D][A][B]=x×x(2−x)×(3−x)=k/2