The equilibrium constant at 298 K for a reaction A + B leftharpoons 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 :

Question

The equilibrium constant at 298 K for a reaction A + B leftharpoons 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 : (A) 0.182 (B) 0.818 (C) 1.818 (D) 1.182

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/915b959b9ad4001f97cce744a26b34e1-.png" /> K C =([ C ][ D ]/[ A ][ B ])=((1+ x )2/(1- x )2)=100 ⇒ (1+ x /1- x )=10 ⇒ 1+x=10-10 x ⇒ 11 x=9 x=(9/11) [ D ] =1+ x =1+(9/11) =1.818 M

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 $m o l L^{- 1}$ ) will be :

0.182

0.818

1.818

1.182

$K_{C} = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{( 1 + x ) ^{2}}{( 1 - x ) ^{2}} = 100$

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

$\Rightarrow 1 + x = 10 - 10 x$

$\Rightarrow 11 x = 9$

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

$[D] = 1 + x = 1 + \frac{9}{11}$

$= 1.818 M$

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