The equilibrium constant at $298 K$ for a reaction $2 A (s) + 2 B (g) ⇌ C (g) + D (g)$ is $100$ . If the initial concentration of all the species were $1 M$ each, then the equilibrium concentration of $D$ (in mol/lit) will be:

0.42

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1.42

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0.16

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1.32

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Solution

The correct option is B $1.42$
$Here, \begin{matrix} k_{c} = \frac{[C] \cdot [D]}{[B]^{2}} & = \frac{(1 + x)^{2}}{(1 - 2 x)^{2}} = 100 . \Rightarrow & \frac{1 + x}{1 - 2 x} = 10 \Rightarrow x = \frac{3}{7} M \end{matrix}$

$Thus, [D] at equilibrium is (1 + \frac{3}{7}) M = 1.42 M$

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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 $D (in mol/l)$ will be:

Q. The equilibrium constant at 298K for a reaction

A + B ⇌ C + D

is 100. If the initial concentrations of all the four species were 1 M each, then equilibrium concentration of D (in mol/L) 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 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 2A(s)+2B(g)⇌C(g)+D(g) is 100. If the initial concentration of all the species were 1M each, then the equilibrium concentration of D (in mol/lit) will be:

The correct option is B 1.42 Here, kc=[C]⋅[D][B]2=(1+x)2(1−2x)2=100.⇒1+x1−2x=10⇒x=37M Thus, [D] at equilibrium is (1+37)M=1.42 M

The equilibrium constant at $298 K$ for a reaction $2 A (s) + 2 B (g) ⇌ C (g) + D (g)$ is $100$ . If the initial concentration of all the species were $1 M$ each, then the equilibrium concentration of $D$ (in mol/lit) will be:

The correct option is B $1.42$
$Here, \begin{matrix} k_{c} = \frac{[C] \cdot [D]}{[B]^{2}} & = \frac{(1 + x)^{2}}{(1 - 2 x)^{2}} = 100 . \Rightarrow & \frac{1 + x}{1 - 2 x} = 10 \Rightarrow x = \frac{3}{7} M \end{matrix}$

$Thus, [D] at equilibrium is (1 + \frac{3}{7}) M = 1.42 M$