Acetic acid ( CH3COOH ) is partially dimerised to (C H3 C O O H)2 in the vapour phase. At a total pressure of text0 text.200 textatm , acetic acid is text92 text.0 % dimerized at text298 textK . The value of equilibrium constant of dimerization under these conditions is

Question

Acetic acid ( CH3COOH ) is partially dimerised to (C H3 C O O H)2 in the vapour phase. At a total pressure of text0 text.200 textatm , acetic acid is text92 text.0 % dimerized at text298 textK . The value of equilibrium constant of dimerization under these conditions is (A) 57.5 textatm- 1 (B) 9.7 textatm- 1 (C) 97 textatm- 1 (D)  text194 textatm- text1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-nqwphuvxygg3.png" /> According to question; text0 text.46 texta text+ text0 text.08 texta text= text0 text.2 ∴ texta text= (0.2/0.54) mathrmK mathrmP=((0.46 × (0.2/0.54))/(0.08 × (0.2)0.54)2) text=194 textatm- text1

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

$57.5 atm^{- 1}$

$9.7 atm^{- 1}$

$97 atm^{- 1}$

$194 atm^{- 1}$

According to question; $0 .46 a + 0 .08 a = 0 .2$

$∴ a = \frac{0.2}{0.54}$

$K_{P} = \frac{( 0.46 \times \frac{0.2}{0.54} )}{( 0.08 \times \frac{0.2}{0.54} ) ^{2}}$

$=194 atm^{- 1}$

Q. Acetic acid ( CH3​COOH ) is partially dimerised to (CH3​COOH)2​ in the vapour phase. At a total pressure of 0.200atm , acetic acid is 92.0% dimerized at 298K . The value of equilibrium constant of dimerization under these conditions is

57.5atm−1

9.7atm−1

97atm−1

194atm−1