MCQ
In reaction $A + 2B \rightleftharpoons 2C + D$, initial concentration of $B$ was $1.5$ times of $[A]$, but at equilibrium the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is
- ✓$4$
- B$6$
- C$12$
- D$8$
✓
Answer
Correct option: A.
$4$
a
$A+2 B \quad \rightleftharpoons \quad 2 C$
$A+2 B \quad \rightleftharpoons \quad 2 C$
$\mathrm{t}=0$
$\mathrm{t}=\mathrm{t}_{\mathrm{eqm}} \quad \mathrm{a}_{\mathrm{o}}$
$a_{0}-x \quad 1.5 a_{0}$
$1.5 a_{0}-2 x$
$2 \mathrm{x}$
$x$
At equilibrium $[\mathrm{A}]=[\mathrm{B}]$
$a_{0}-x=1.5 a_{0}-2 x \Rightarrow x=0.5 a_{0}$
$[\mathrm{C}]=2 \times 0.5 \mathrm{a}_{\mathrm{o}}=\mathrm{a}_{\mathrm{o}},[\mathrm{D}]=[\mathrm{A}]=[\mathrm{B}]=0.5 \mathrm{a}_{\mathrm{o}}$
$\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{C}]^{2}[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]^{2}}=\frac{\left(\mathrm{a}_{0}\right)^{2}\left(0.5 \mathrm{a}_{0}\right)}{\left(0.5 \mathrm{a}_{0}\right)\left(0.5 \mathrm{a}_{0}\right)^{2}}=4$
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