In the reaction A + 2B rightleftharpoons 2C + | vedclass

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(A) The reaction is $A + 2B \rightleftharpoons 2C + D$.Let the initial concentration of $A$ be $a_0$. Then the initial concentration of $B$ is $1.5a_0

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The reaction is $A + 2B ightleftharpoons 2C + D$.Let the initial concentration of $A$ be $a_0$. Then the initial concentration of $B$ is $1.5a_0$.At $t=0$: $[A] = a_0$,$[B] = 1.5a_0$,$[C] = 0$,$[D] = 0$.At equilibrium $(t=t_{eqm})$: $[A] = a_0 - x$,$[B] = 1.5a_0 - 2x$,$[C] = 2x$,$[D] = x$.Given that at equilibrium $[A] = [B]$:$a_0 - x = 1.5a_0 - 2x \Rightarrow x = 0.5a_0$.Substituting $x$ into the equilibrium concentrations:$[A] = a_0 - 0.5a_0 = 0.5a_0$$[B] = 1.5a_0 - 2(0.5a_0) = 0.5a_0$$[C] = 2(0.5a_0) = a_0$$[D] = x = 0.5a_0$The equilibrium constant $K_C$ is given by:$K_C = \frac{[C]^2 [D]}{[A][B]^2} = \frac{(a_0)^2 (0.5a_0)}{(0.5a_0)(0.5a_0)^2} = \frac{a_0^2 	imes 0.5a_0}{0.5a_0 	imes 0.25a_0^2} = \frac{1}{0.25} = 4$.

In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:

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