For A + B ⇌ C + D , if initially the concentration of A and B are equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of C + D ⇌ A + B reaction ?

Correct answer: C. (1)/(4) x = 2 (a - x) 3x = 2a x = (2a)/(3) K _ c x ⋅ x (a – x)(a – x) = 4a^ 2 9 × (a)/(3) × (a)/(3) =4 C+D ⇌ A+B; K _ c = (1)/(4)

NTA Abhyas 2020

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

For A + B ⇌ C + D , if initially the concentration of A and B are...

For

A + B ⇌ C + D

, if initially the concentration of

\text{A}

and

\text{B}

are equal but at equilibrium concentration of

\text{D}

will be twice of that of

\text{A}

then what will be the equilibrium constant of

C + D ⇌ A + B

reaction ? For A + B ⇌ C + D , if initially the concentration of A and B are equal but at equilibrium concentration of D will be

C. $\dfrac{1}{4}$

\text{x} = 2 \left(\text{a} - \text{x}\right)<br />3x = 2a

x = \dfrac{2a}{3}<br />

{K}_{c}

\dfrac{\text{x} ⋅ \text{x}}{\left(\text{a} – \text{x}\right)\left(\text{a} – \text{x}\right)} = \dfrac{4a^{2}}{9 \times \dfrac{a}{3} \times \dfrac{a}{3}} =4

\text{C}\text{+}\text{D} ⇌ \text{A}\text{+}\text{B;}{K}_{c} = \dfrac{1}{4}