Question

If two gases $A{B}_2$ and $B_{2}C$ are mixed the following equilibria are readily established
$A{B}_2\left(g\right)+B_{2}C\left(g\right)\to A{B}_3\left(g\right)+BC\left(g\right)$
$BC\left(g\right)+B_{2}C\left(g\right)\to B_3{C}_2\left(g\right)$
If the reaction is started only with $A{B}_2$ with $B_{2}C$, then which of the following is necessarily true at equilibrium:

• A. $[A{B}_3{]}_{eq}=[BC{]}_{eq}$
• B. $[A{B}_3{]}_{eq}>[BC{]}_{eq}$
• C. $[A{B}_3{]}_{eq}>[B_3{C}_2{]}_{eq}$
• D. $[A{B}_2{]}_{eq}=[B_{2}C{]}_{eq}$

Answer 1

Answer: (B), (C)

Let reactions is started with a mole of $A{B}_2$ and b mole of $B_{2}C$
$\Rightarrow A{B}_2\left(g\right)+B_{2}C\left(g\right)\to A{B}_3\left(g\right)+BC\left(g\right)$
a b 0 0
a - x b - x - y x x - y
$BC\left(g\right)+B_{2}C\left(g\right)\to B_2{C}_2\left(g\right)$
x - y b - x - y y As x > y
Clearly $[A{B}_3{]}_{eq}>[B_3{C}_2{]}_{eq}$ and $[A{B}_3{]}_{eq}>[BC{]}_{eq}$