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Standard XII

Chemistry

Derivation of Kp and Kc

In an equilib...

Question

In an equilibrium $A + B ⇌ C + D$ , $A$ and $B$ are mixed in a vessel at temperature $T$ . The initial concentration of $A$ was twice the initial concentration of $B$ . After the equilibrium has reached, concentration of $C$ was thrice the equilibrium concentration of $B$ .Calculate $K_{C}$ .

0.14

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3.6

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1.8

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Solution

The correct option is D 1.8
Let the inital concentration of $B = a$ mol litre $^{- 1}$ and thus, $[A] = 2 a$ (given)

$A + B ⇌ C + D$
Initial concentration 2a a 0 0
Eqm concentration $(2 a - x) (a - x) x x$

The expression for the equilibrium constant is as given blow.
$K = \frac{x^{2}}{(2 a - x) (a - x)}$ .......(i)

$∵$ At equilibrium $[C] = 3 [B]$

or $x = 3 (a - x)$

$∴$ $x = \frac{3 a}{4}$ .......(ii)
By equations (i) and (ii) $K = \frac{{(3 a / 4)}^{2}}{[2 a - (3 a / 4)] [a - 3 a / 4]} = 1.8$

So, the correct answer is 1.8

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In an equilibrium A+B⇌C+D, A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reached, concentration of C was thrice the equilibrium concentration of B.Calculate KC.

In an equilibrium $A + B ⇌ C + D$ , $A$ and $B$ are mixed in a vessel at temperature $T$ . The initial concentration of $A$ was twice the initial concentration of $B$ . After the equilibrium has reached, concentration of $C$ was thrice the equilibrium concentration of $B$ .Calculate $K_{C}$ .