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Equilibrium Constant and Standard Free Energy Change

X2 + Y2⇌ 2XY ...

Question

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

(\frac{1}{3}, 0.3), (\frac{2}{3}, 0.3)

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(\frac{1}{3}, 0.6), (\frac{2}{3}, 0.6)

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(1, -0.3), (2, -0.3)

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(1, -0.6), (2, -0.6)

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Solution

The correct option is A $(\frac{1}{3}, 0.3), (\frac{2}{3}, 0.3)$
$\begin{matrix} Total volume used in reaction \begin{matrix} = 1 + 2 = 3 L [x_{2}] = \frac{1}{3} & [y_{2}] = \frac{2}{3} \end{matrix} \end{matrix}$
$x_{2} + y_{2} ⇌ 2 x y$
$t = 0 \frac{1}{3} \frac{2}{3}$
$t = t \frac{1}{3} - x \frac{2}{3} - x 2 x$
$\begin{matrix} as [x y] & = 0.6 \Rightarrow & 2 x = 0.6 \Rightarrow x & = 0.3 ∴ [x_{2}] = \frac{1}{3} - x = \frac{1}{3} & - 0.3 = 0.033 and, [y_{2}] = \frac{2}{3} - x & = \frac{2}{3} - 0.3 = 0.367 \end{matrix}$

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X_{2} + Y_{2} ⇌ 2 X Y

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

The correct option is A (13,0.3),(23,0.3) Total volume used in reaction =1+2=3L [ x2]=13 & [y2]=23x2+y2⇌2xyt=01323t=t13−x23−x2xas[xy]=0.6⇒2x=0.6⇒x=0.3∴[x2]=13−x=13−0.3=0.033 and, [y2]=23−x=23−0.3=0.367