For the reaction. A ( l ) → 2 B ( g ) ΔU = 2. 1k Cal , ΔS = 20 Cal K − 1 at 300 K Hence Δ G in k c a l is?

Δ H = Δ U + Δ n g R T = 2.1 × 1 0 3 + 2 2 ( 300 ) = 33 00 Cal Δ G = Δ H - T Δ S = 33 00 – 300 20 = – 27 00 Cal = − 2.7 K C a l

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ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2020

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For the reaction.
$A (l) \to 2 B (g)$
$ΔU = 2. 1k Cal, ΔS = 20 Cal K^{- 1} at 300 K$
Hence $Δ G$ in $k c a l$ is?

Answer:

Solution:

$Δ H = Δ U + Δ n g R T$
$= 2.1 \times 1 0^{3} + 2 (2) (300) = 3300 Cal$
$Δ G = Δ H - T Δ S$
$= 3300 - 300 (20) = - 2700$ $Cal$
$= - 2.7$ $K C a l$

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

⏱ 2mℹ️ Source: PYQ_2020

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