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?

For the reaction.

A(l)→2B(g)

ΔU=2.1k Cal, ΔS=20 Cal  K  −1  at 300 K

Hence ΔG in kcal is?
  1. 2.7
  2. – 2.7
  3. 5.4
  4. 1.35

Solution

ΔH=ΔU+ΔngRT
=2.1×103+22(300)=3300 Cal
ΔG=ΔH-TΔS
=3300–30020=–2700 Cal
=−2.7 K Cal .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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