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$8 \mathrm{~mol}$ of $A B_3(\mathrm{~g})$ are introduced into a $1.0 \mathrm{dm}^3$ vessel. If it dissociates as
$2 \mathrm{AB}_3(g) \rightleftharpoons A_2(g)+3 B_2(g)$. At equilibrium, $2 \mathrm{~mol}$ of $A_2$ are found to be present. The equilibrium constant of this reaction is
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$2 \mathrm{AB}_3(g) \rightleftharpoons A_2(g)+3 B_2(g)$. At equilibrium, $2 \mathrm{~mol}$ of $A_2$ are found to be present. The equilibrium constant of this reaction is
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The correct answer is:
27
27
$2 A B_3(g) \leftrightharpoons A_2(g)+3 B_2(g)$
at $t=0 \quad 8 \quad 0 \quad 0$
at eq. $(8-2 \times 2) \quad 2 \quad 3 \times 2$ $=4 \quad 2 \quad 6$
now $K_C=\frac{\left[A_2\right]\left[B_2\right]^3}{\left[A B_3\right]^2}=\frac{2 / 1 \times[6 / 1]^3}{[4 / 1]^2}=27$
at $t=0 \quad 8 \quad 0 \quad 0$
at eq. $(8-2 \times 2) \quad 2 \quad 3 \times 2$ $=4 \quad 2 \quad 6$
now $K_C=\frac{\left[A_2\right]\left[B_2\right]^3}{\left[A B_3\right]^2}=\frac{2 / 1 \times[6 / 1]^3}{[4 / 1]^2}=27$
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