Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2020

ChemistryEasysingle choice

At $25 ° C$ the enthalpy change, for the ionization of trichloroacetic acid is $+ 6.3 k J m o l^{- 1}$ and the entropy change, is $+ 0.0084 k J m o l^{- 1} K^{- 1}$ . Then $p K a$ of tricholoro acetic acid is

Options:

Answer:

Solution:

$Δ G ° = Δ H ° - T Δ S °$
$Δ G ° = 6.3 - (298 \times 0.0084) = 3.8 K J . m o l^{- 1}$
$∴ (Δ G^{o} = 2.303 RTlnK)$
$l o g K e q = l o g K a = - p K a = \frac{- Δ G °}{2.303 R T}$
$- p K a = \frac{- Δ G °}{2.303 R T} = \frac{3.8}{2.303 \times 0.008314 \times 298}$
$= 0.66$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2020

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