For the reaction ; A.(l) → 2B(g) Δ U = 2.1 kcal, Δ S = 20 cal K-1 at 300 K. Hence ΔG in kcal is.

Question

For the reaction ; A.(l) → 2B(g) Δ U = 2.1 kcal, Δ S = 20 cal K-1 at 300 K. Hence ΔG in kcal is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A(ℓ) → 2B(g) Δ U=2.1 kcal, Δ s=20 cal K-1 at 300 k Δ H=Δ U+Δ ng RT Δ G=Δ H-TΔ S Δ G=Δ U+Δ ngRT-TΔ S =2.1+(2×2×300/1000)-(300×20/1000) (R=2 cal k-1mol-1) =2.1+1.2-6=-2.70 Kcal/mol

Q. For the reaction ; A.(l)→2B(g) ΔU=2.1kcal,ΔS=20calK−1at300K. Hence ΔG in kcal is__________.

A(ℓ)→2B(g) ΔU=2.1kcal,Δs=20calK−1at300k ΔH=ΔU+Δng​RT<br/>ΔG=ΔH−TΔS ΔG=ΔU+Δng​RT−TΔS =2.1+10002×2×300​−1000300×20​ (R=2calk−1mol−1) =2.1+1.2−6=−2.70Kcal/mol

Q. For the reaction ;
$A . (l) \to 2 B (g)$
$Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1} a t 300 K .$
Hence $Δ$ G in kcal is__________.