For the reaction, 2 A (g) + B (g) ⇌ C (g) + D (g) ; K _ c = 1 0 ^ 1 2 . If the initial moles of A,B,C and D are 2, 1, 7 and 3 moles respectively in 1L vessel, what is the equilibrium concentration of A?

NTA Abhyas 2020

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

For the reaction, 2 A (g) + B (g) ⇌ C (g) + D (g) ; K _ c = 1 0 ^ 1...

For the reaction,

2 \text{A} \left(\text{g}\right) + \text{B} \left(\text{g}\right) ⇌ \text{C} \left(\text{g}\right) + \text{D} \left(\text{g}\right) {; K}_{c} = 1 {0}^{1 2}

. If the initial moles of

\text{A,}\text{B,}\text{C}

and

\text{D}

are

2, 1, 7

and

3

moles respectively in

\text{1}\text{L}

vessel, what is the equilibrium concentration of

\text{A?}

A. $4 \times 1 {0}^{- 4}$

\begin{cases} & 2 \text{A} \left(\text{g}\right) & + & \text{B} \left(\text{g}\right) & ⇌ & \text{C} \left(\text{g}\right) & + & \text{D} \left(\text{g}\right) \\ \text{Initial moles} & 2 & & 1 & & 7 & & 3 \\ \text{At equilibrium} & 2 - 2 \text{x} & & 1 - \text{x} & & 7 + \text{x} & & 3 + \text{x}\end{cases}

∴

Due to very high value of

\text{K,}

we can assume that reactant almost gets converted into products, so

1 - \text{x} = \text{y}

2 - 2 \text{x} = 2 \text{y} ⇒ \text{x} ∼ 1

∴ 1 {0}^{1 2} = \dfrac{\left[\text{C}\right] \left[\text{D}\right]}{{\left[\text{A}\right]}^{2} \left[\text{B}\right]}

1 {0}^{1 2} = \dfrac{8 \times 4}{{\left(2 \text{y}\right)}^{2} \left(\text{y}\right)}

y^{3} = 8 \times 1 {0}^{- 1 2}

⇒ \text{y} = 2 \times 1 {0}^{- 4}

∴

Equilibrium concentration of

\text{A} = 2 \text{y}

= 2 \times 2 \times 1 {0}^{- 4} = 4 \times 1 {0}^{- 4}