When the heat of a reaction at constant pressure is - 25 × 10 - 3 cal and entropy change for the reaction is 7 . 4 cal deg - 1 , it is predicted that the reaction at 25 0 C is

Heat at constant pressure means enthalpy. i.e., ΔG = ΔH - TΔS ΔH = 2 . 5 × 10 3 cal ΔS = 7 . 4 caldeg - 1 T = 298 K ΔG = ΔH - TΔS = - 25 × 10 3 - 298 × 7 . 4 = - 4705 . 2 cal As the value of ΔG is −4705.2 cal, which is a negative value, so this reaction will be spontaneous. Hence, the process is spontaneous.

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ChemistryThermodynamics - IICalculation of DSEasy2 minPYQ_2018

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When the heat of a reaction at constant pressure is $- 25 \times 10^{- 3} cal$ and entropy change for the reaction is $7.4 cal {deg}^{- 1}$ , it is predicted that the reaction at $25^{0} C$ is

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Solution:

Heat at constant pressure means enthalpy. i.e., $ΔG = ΔH - TΔS$

$ΔH = 2.5 \times 10^{3} cal$

$ΔS = 7.4 {caldeg}^{- 1}$

$T = 298 K$

$ΔG = ΔH - TΔS$

$= - 25 \times 10^{3} - 298 \times 7.4$

$= - 4705.2 cal$

As the value of ΔG is −4705.2 cal, which is a negative value, so this reaction will be spontaneous.

Hence, the process is spontaneous.

Stream:BITSATSubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS

⏱ 2mℹ️ Source: PYQ_2018

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