The equilibrium constant at 298K 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 $m o l L^{- 1}$ ) will be.

0.818

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1.818

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1.182

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0.182

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Solution

The correct option is B 1.818
$A + B \to C + D 1111 : i n i t i a l l y (1 - x) (1 - x) (1 + x) (1 + x) : a t e q u i l i b r i u m$

$K c = {(\frac{1 + x}{1 - x})}^{2}$
$K c = {(\frac{1 + x}{1 - x})}^{2} = 100$

$\frac{1 + x}{1 - x} = 10$

$x = \frac{9}{11}$

Moles of $D = 1 + x = 1 + \frac{9}{11} = \frac{20}{11} = 1.818$

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The equilibrium constant at 298K 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 molL−1) will be.

The correct option is B 1.818A + B→ C + D1 1 1 1 :initially(1−x)(1−x) (1+x) (1+x):at equilibrium Kc=(1+x1−x)2 Kc=(1+x1−x)2=100 1+x1−x=10 x=911 Moles of D=1+x=1+911=2011=1.818

The equilibrium constant at 298K 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 $m o l L^{- 1}$ ) will be.