Question

$1$ mole of $CH _{3} COOH$ and $1$ mole of $C _{2} H _{5} OH$ reacts to produce $\frac{2}{3}$ mole of $CH _{3} COOC _{2} H _{5}$. The equilibrium constant is

• A. 2
• B. + 2
• C. -4
• D. + 4

Answer 1

Answer: (D)

Let the total volume $= V L$
$\therefore \left[ CH _{3} OOH \right]=\frac{1}{3} V mol / L$
$\left[ C _{2} H _{5} OH \right]=\frac{1}{3} V mol / L$
$\left[ CH _{3} COOC _{2} H _{5}\right]=\frac{2}{3} V mol / L$
$\left[ H _{2} O \right]=\frac{2}{3} V mol / L$
$\therefore K =\frac{\frac{2}{3} V \times \frac{2}{3} V }{\frac{1}{3} V \times \frac{1}{3} V }=4$