Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIAdiabatic Process/PolytropicMedium2 minPYQ_2020

ChemistryMediumnumerical

The value of ${log}_{10} K$ for reaction, $A ⇌ B ;$ is

(Given : $Δ_{r} H_{298 K}^{0} = - 54.07 k J m o l^{- 1}, Δ_{r} S_{288 K}^{0} = 10 J K^{- 1} m o l^{- 1}$ and $R = 8.31 J K^{- 1} m o l^{- 1}, 2.303 \times 8.314 \times 298 = 5705$ )

Answer:

10.00

Solution:

$log K = \frac{{ΔS}^{o}}{2 .303 R} - \frac{{ΔH}^{o}}{2 .303 RT}$
$= \frac{10}{2.303 \times 8.314} - \frac{(- 54.07 \times 10^{3})}{2.303 \times 8.314 \times 298}$
$= \frac{10 \times 298}{5705} + \frac{54070}{5705}$
$= \frac{57050}{5705} = 10$

Stream:NTA_ABHYASSubject:ChemistryTopic:Thermodynamics - IISubtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...