The gas A_ 2 in the left flask allowed to react with gas B_ 2 present in right flask as, A _ 2 (g) + B _ 2 (g) ⇌ 2 AB (g) ; K_ c =4 at 27^ o C. What is the concentration of AB when equilibrium is established?

Correct answer: C. 0.66 M K _ c = [AB]2 [ A _ 2 ][ B _ 2 ] K _ c = 4x^ 2 (2 - x)(4 - x) ⇒ x = (3 2)/(2 4) = 1.33 mole AB(g) = 2 × 1.33 4 = 0.66 M

NTA Abhyas 2020

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The gas A_ 2 in the left flask allowed to react with gas B_ 2...

The gas A_{2} in the left flask allowed to react with gas B_{2} present in right flask as,

{A}_{2} \left(\text{g}\right) + {B}_{2} \left(\text{g}\right)

⇌ 2

\text{AB} \left(\text{g}\right)

; K_{c}=4 at 27^{o}C. What is the concentration of AB when equilibrium is established?

C. 0.66 M

{K}_{c} = \dfrac{\left[AB\right]2}{\left[{A}_{2}\right]\left[{B}_{2}\right]}

{K}_{c} = \dfrac{4x^{2}}{\left(2 - \text{x}\right)\left(4 - \text{x}\right)}

⇒ \text{x} = \dfrac{3 2}{2 4} = \text{1.33 mole}

\text{AB}\left(\text{g}\right) = \dfrac{2 \times \text{1}\text{.33}}{4} = \text{0}\text{.66 M}