Consider the following reaction taking place in $1 ext{ L}$ capacity container at $300 ext{ K}$:
$ ext{A} +

Question

Consider the following reaction taking place in $1	ext{ L}$ capacity container at $300	ext{ K}$:
$	ext{A} + 	ext{B} ightleftharpoons 	ext{C} + 	ext{D}$
If one mole each of A and B are present initially and at equilibrium $0.7	ext{ mol}$ of C is formed, then the equilibrium constant ($K_c$) for the reaction is:

Accepted Answer

For the given reversible reaction:
$	ext{A} + 	ext{B} ightleftharpoons 	ext{C} + 	ext{D}$

Initial moles: 
$n_A = 1$
$n_B = 1$
$n_C = 0$
$n_D = 0$

Let $x$ be the number of moles of C formed at equilibrium. The moles of each species at equilibrium will be:
$n_A = 1 - x$
$n_B = 1 - x$
$n_C = x$
$n_D = x$

We are given that $0.7	ext{ mol}$ of C is formed at equilibrium, so $x = 0.7$.
Since the volume of the container is $1	ext{ L}$, the molar concentrations at equilibrium are equal to the number of moles :
$[	ext{A}] = 1 - 0.7 = 0.3	ext{ M}$
$[	ext{B}] = 1 - 0.7 = 0.3	ext{ M}$
$[	ext{C}] = 0.7	ext{ M}$
$[	ext{D}] = 0.7	ext{ M}$

The equilibrium constant expression ($K_c$) is:
$K_c = \frac{[	ext{C}][	ext{D}]}{[	ext{A}][	ext{B}]}$

Substituting the equilibrium concentrations into the expression:
$K_c = \frac{0.7 	imes 0.7}{0.3 	imes 0.3} = \frac{0.49}{0.09} = \frac{49}{9} \approx 5.44$

Therefore, the equilibrium constant ($K_c$) is approximately $5.4$.

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