A_{(g)} + 3B_{(g)} rightleftharpoons 4C_{(g)} | vedclass

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(C) The reaction is $A_{(g)} + 3B_{(g)} \rightleftharpoons 4C_{(g)}$.Let the initial concentration of $A$ and $B$ be $a \ mol/L$.Initial concentration

Vedclass · Accepted Answer

$8$

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(C) The reaction is $A_{(g)} + 3B_{(g)} \rightleftharpoons 4C_{(g)}$.Let the initial concentration of $A$ and $B$ be $a \ mol/L$.Initial concentrations: $[A] = a, [B] = a, [C] = 0$.At equilibrium,let the concentration of $A$ reacted be $x$.Equilibrium concentrations: $[A] = a - x$,$[B] = a - 3x$,$[C] = 4x$.Given that at equilibrium,$[A] = [C]$,so $a - x = 4x$,which implies $a = 5x$.Substituting $a = 5x$ into the equilibrium concentrations:$[A] = 5x - x = 4x$$[B] = 5x - 3x = 2x$$[C] = 4x$$K_c = \frac{[C]^4}{[A][B]^3} = \frac{(4x)^4}{(4x)(2x)^3} = \frac{256x^4}{(4x)(8x^3)} = \frac{256x^4}{32x^4} = 8$.

$A_{(g)} + 3B_{(g)} \rightleftharpoons 4C_{(g)}$. The starting concentrations of $A$ and $B$ are equal. At equilibrium,the concentrations of $A$ and $C$ are the same. The value of $K_c$ is:

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