For the reaction, X2 O 4(l) longrightarrow 2 XO 2(g), Δ U=2.1 kcal , Δ S=20 cal K -1 at 300 K. Hence, Δ G is

Question

For the reaction, X2 O 4(l) longrightarrow 2 XO 2(g), Δ U=2.1 kcal , Δ S=20 cal K -1 at 300 K. Hence, Δ G is (A) 2.7 kcal (B) -2.7 kcal (C) 9.3 kcal (D) -9.3 kcal

Tardigrade Admin · Accepted Answer

As we know that, Δ H=Δ U+Δ ng R T where, Δ U= change in internal energy Δ ng= number of moles of gaseous products - number of moles of gaseous reactants =2-0=2 R= gas constant =2 cal But, Δ U=2.1 kcal =2.1 × 103 cal [∵ 1 kcal =103 cal ] ∴ Δ H=(2.1 × 103)+(2 × 2 × 300)=3300 cal Now, Δ G=Δ H-T Δ S ⇒ Δ G=(3300)-(300 × 20) ⇒ Δ G=-2700 cal ∴ Δ G=-2.7 kcal

Q. For the reaction, $X_{2} O_{4} (l) ⟶ 2 X O_{2} (g)$ , $Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1}$ at $300 K$ . Hence, $Δ G$ is

$2.7 k c a l$

$- 2.7 k c a l$

$9.3 k c a l$

$- 9.3 k c a l$

Q. For the reaction, X2​O4​(l)⟶2XO2​(g), ΔU=2.1kcal,ΔS=20calK−1 at 300K. Hence, ΔG is

2.7kcal

−2.7kcal

9.3kcal

−9.3kcal

Q. For the reaction, $X_{2} O_{4} (l) ⟶ 2 X O_{2} (g)$ , $Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1}$ at $300 K$ . Hence, $Δ G$ is

$2.7 k c a l$

$- 2.7 k c a l$

$9.3 k c a l$

$- 9.3 k c a l$