Acetic acid forms a dimer in the gas phaseThe dimer is held together by two hydrogen bonds with a total strength of 66-5 kJ per mole of dimer- At 25- C- the equilibrium constant for the dimerisation is 1-3 - 103 pressure in atm- What is magnitude S 0 for the reaction- Assume that H does not vary with temperature-

2CH_3COOH(g) → (CH_3COOH)_2(g) G^o 2.303RT log K G^o = 2.303 × 8.314 × 298 × log(1.3 × 10^3) = 17767.69J = 17.77kJ Again, nbsp; G^0 = H^0 T S^0 17. ...

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Byju's Answer

Standard XII

Chemistry

Entropy for a Reversible Process

Acetic acid f...

Question

Acetic acid forms a dimer in the gas phase
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer. At $25^{o} C,$ the equilibrium constant for the dimerisation is $1.3 \times 10^{3}$ (pressure in atm). What is magnitude $△ S^{0}$ for the reaction? Assume that $△ H$ does not vary with temperature.

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Solution

$2 C H_{3} C O O H (g) \to (C H_{3} C O O H)_{2} (g)$
$△ G^{o} - - 2.303 R T l o g K$
$△ G^{o} = - 2.303 \times 8.314 \times 298 \times l o g (1.3 \times 10^{3}) = - 17767.69 J = - 17.77 k J$

Again,
$△ G^{0} = △ H^{0} - T △ S^{0}$
$- 17.77 = - 66.5 - 298 \times △ S^{0}$
$△ S^{0}$ = $(\frac{- 66.5 + 17.77}{298})$
$△ S^{0}$ = $(\frac{- 48.73}{298}) = - 0.163 k J m o l^{- 1} K^{- 1}$
Therefore, $△ S^{o} = - 0.163 k J m o l^{- 1} K^{- 1}$

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2CH3COOH(g)→(CH3COOH)2(g) △Go−−2.303RT log K △Go=−2.303×8.314×298×log(1.3×103)=−17767.69J=−17.77kJ Again, △G0=△H0−T△S0 −17.77=−66.5−298×△S0 △S0 = (−66.5+17.77298) △S0 = (−48.73298)=−0.163 kJ mol−1 K−1 Therefore, △So=−0.163kJ mol−1 K−1