At a definite temperature, the equilibrium constant for a reaction, A + B ⇌ 2 C & , was found to be 81. Starting with 1 mole A and 1 mole B, the mole fraction of C at equilibrium is :

Correct answer: A. (9)/(1 1) cases A & + & B & ⇌ & 2 C 1 & & 1 & & 0 1 - x & & 1 - x & & 2 x cases ∴ 8 1 = 4 x^ 2 (1 - x) ^ 2 or 2 x (1 - x) = 9 ∴ x = (9)/(1 1) Thus, cases A & + & B & ⇌ & 2 C (2)/(1 1) & & (2)/(1 1) & & (18)/(1 1) cases ( Total mole = (2)/(1 1) + (2)/(1 1) + (18)/(1 1) = (22)/(1 1) ) ∴ Mole fraction of C= ((18)/(1 1))/((2 2)/(1 1)) = (9)/(1 1)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

At a definite temperature, the equilibrium constant for a reaction,...

At a definite temperature, the equilibrium constant for a reaction,

\text{A} + \text{B} ⇌ 2 \text{C}

& , was found to be 81. Starting with 1 mole A and 1 mole B, the mole fraction of C at equilibrium is :

Options

A $\dfrac{9}{1 1}$
B $\dfrac{1}{1 1}$
C $\dfrac{2}{1 1}$
D $\dfrac{7}{1 1}$

Correct answer

A. $\dfrac{9}{1 1}$

Step-by-step solution

\begin{cases}\text{A} & + & \text{B} & ⇌ & 2 \text{C} \\ 1 & & 1 & & 0 \\ 1 - \text{x} & & 1 - \text{x} & & 2 \text{x}\end{cases}

∴ 8 1 = \dfrac{4 x^{2}}{{\left(1 - \text{x}\right)}^{2}}

\dfrac{2 \text{x}}{\left(1 - \text{x}\right)} = 9

∴ \text{x} = \dfrac{9}{1 1}

Thus,

\begin{cases}\text{A} & + & \text{B} & ⇌ & 2 \text{C} \\ \dfrac{2}{1 1} & & \dfrac{2}{1 1} & & \dfrac{18}{1 1}\end{cases}

(

Total mole

= \dfrac{2}{1 1} + \dfrac{2}{1 1} + \dfrac{18}{1 1} = \dfrac{22}{1 1} )

∴

Mole fraction of

C= \dfrac{\dfrac{18}{1 1}}{\dfrac{2 2}{1 1}} = \dfrac{9}{1 1}

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