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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

X2+Y2⇌ 2XY re...

Question

$X_{2} + Y_{2} ⇌ 2 X Y$ reaction was studied at a certain temperature. In the beginning $1$ mole of $X_{2}$ was taken in a one litre flask and $2$ moles of $Y_{2}$ was taken in another $2$ litre flask. What is the equilibrium concentration of $X_{2}$ and $Y_{2}$ , respectively?(Given equilibrium concentration of $[X Y] = 0.6 {m o l}^{- 1}$ )

(\frac{1}{3} - 0.3)

and

(\frac{2}{3} - 0.3)

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(\frac{1}{3} - 0.6)

and

(\frac{2}{3} - 0.6)

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(1-0.3) and (2-0.3)

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(1-0.6) and (2-0.6)

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Solution

The correct option is A $(\frac{1}{3} - 0.3)$ and $(\frac{2}{3} - 0.3)$
$X_{2}$ $Y_{2}$ $⇌$ $2 X Y$
At t=0 $\frac{1}{3}$ $\frac{2}{3}$ $0$
At equilibrium $\frac{1}{3}$ $- x$ $\frac{2}{3}$ $- x$ $2 x$
But $2 x = 0.6$
$x = 0.3$
Equilibrium concentration of $X_{2}$ = $\frac{1}{3}$ - $x$ = $\frac{1}{3}$ - $0.3$
Equilibrium concentration of $Y_{2}$ = $\frac{2}{3}$ - $x$ = $\frac{2}{3}$ - $0.3$

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X_{2} + Y_{2} ⇌ 2 X Y

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X2+Y2⇌2XY reaction was studied at a certain temperature. In the beginning 1 mole of X2 was taken in a one litre flask and 2 moles of Y2 was taken in another 2 litre flask. What is the equilibrium concentration of X2 and Y2, respectively?(Given equilibrium concentration of [XY]=0.6mol−1)

The correct option is A (13−0.3) and (23−0.3) X2Y2⇌2XYAt t=0 13 23 0At equilibrium 13−x 23−x 2xBut 2x=0.6x=0.3Equilibrium concentration ofX2=13-x=13-0.3Equilibrium concentration ofY2=23-x=23-0.3