Question

Acetic acid ( $CH_{3}COOH$ ) is partially dimerised to $\left(C H_{3} C O O H\right)_{2}$ in the vapour phase. At a total pressure of $\text{0} \text{.200} \, \text{atm}$ , acetic acid is $\text{92} \text{.0\%}$ dimerized at $\text{298} \, \text{K}$ . The value of equilibrium constant of dimerization under these conditions is

• A. $57.5 \, \text{atm}^{- 1}$
• B. $9.7 \, \text{atm}^{- 1}$
• C. $97 \, \text{atm}^{- 1}$
• D. $\text{194} \, \text{atm}^{- \text{1}}$

Answer 1

Answer: (D)

According to question; $\text{0} \text{.46} \, \text{a} \, \text{+} \, \text{0} \text{.08} \, \text{a} \, \text{=} \, \text{0} \text{.2}$

$\therefore \, \text{a} \, \text{=} \, \frac{0.2}{0.54}$

$\mathrm{K}_{\mathrm{P}}=\frac{\left(0.46 \times \frac{0.2}{0.54}\right)}{\left(0.08 \times \frac{0.2}{0.54}\right)^2}$

$\text{=194} \, \text{atm}^{- \text{1}}$