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

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(B) The reaction is $A + 2B \rightleftharpoons 2C + D$. Initial concentrations: $[A] = a$,$[B] = 1.5a$,$[C] = 0$,$[D] = 0$. At equilibrium,let the con

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$a/2$

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(B) The reaction is $A + 2B \rightleftharpoons 2C + D$. Initial concentrations: $[A] = a$,$[B] = 1.5a$,$[C] = 0$,$[D] = 0$. At equilibrium,let the concentration of $D$ be $x$. According to stoichiometry,$[A]_{eq} = a - x$,$[B]_{eq} = 1.5a - 2x$,$[C]_{eq} = 2x$,$[D]_{eq} = x$. Given that at equilibrium $[A] = [D]$,so $a - x = x$,which implies $2x = a$ or $x = a/2$. Substituting $x = a/2$ into the expression for $[B]_{eq}$: $[B]_{eq} = 1.5a - 2(a/2) = 1.5a - a = 0.5a = a/2$. Species$A + 2B \rightleftharpoons 2C + D$Initial concentration$a, 1.5a, 0, 0$Equilibrium concentration$(a - x), (1.5a - 2x), 2x, x$

For the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $A$ is $a$ and the initial concentration of $B$ is $1.5$ times that of $A$. If the concentrations of $A$ and $D$ are equal at equilibrium,what will be the concentration of $B$ at equilibrium?

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