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Characteristics of Equilibrium Constant

The equilibri...

Question

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:

1.818

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1.182

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0.182

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0.818

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Solution

The correct option is D $0.182$
$A + B ⇌ C + D$
$t = 0$ $1$ $1$ $1$ $1$
$t_{e q}$ $1 - x$ $1 - x$ $1 + x$ $1 + x$
$k = \frac{(1 + x)^{2}}{(1 - x)^{2}} = 100$
$\Rightarrow \frac{1 + x}{1 - x} = 10$
$\Rightarrow 1 + x = 10 - 10 x$
$\Rightarrow x = \frac{9}{11} = 0.818$
$∴ [D] = 1.818$ .

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Similar questions

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 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:

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:

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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:

The correct option is D 0.182 A+B⇌C+Dt=0 1 1 1 1teq 1−x 1−x 1+x 1+xk=(1+x)2(1−x)2=100⇒1+x1−x=10⇒1+x=10−10x⇒x=911=0.818∴[D]=1.818.

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:

The correct option is D $0.182$
$A + B ⇌ C + D$
$t = 0$ $1$ $1$ $1$ $1$
$t_{e q}$ $1 - x$ $1 - x$ $1 + x$ $1 + x$
$k = \frac{(1 + x)^{2}}{(1 - x)^{2}} = 100$
$\Rightarrow \frac{1 + x}{1 - x} = 10$
$\Rightarrow 1 + x = 10 - 10 x$
$\Rightarrow x = \frac{9}{11} = 0.818$
$∴ [D] = 1.818$ .