For the reaction: X2 O4(l) - 2XO2(g) Δ U = 2.1 k cal , Δ S = 20 cal k-1 at 300 K Hence Δ G is: -

Question

For the reaction: X2 O4(l) -> 2XO2(g) Δ U = 2.1 k cal , Δ S = 20 cal k-1 at 300 K Hence Δ G is: - (A) 2.7 k cal (B) - 2.7 k cal (C) 9.3 k cal (D) - 9.3 k cal

Tardigrade Admin · Accepted Answer

The change in Gibbs Free energy is given by

Δ H = Δ U + Δ n_{g} RT

where

Δ H

is the enthalpy of the reaction

Δ S

is the entropy of the reaction and

Δ U

is the change in internal energy

Δ n_{g}

is the (number of gaseous moles in product) - (number of gaseous moles in reactant )

= 2 - 0 = 2

R

is the gas constant

= 2

cal But,

Δ H = (2.1 \times 1 0^{3}) + (2 \times 2 \times 300) = 3300 c a l

Hence,

Δ G = Δ H = T Δ S

Δ G = 3300 - (300 \times 20)

Δ G = - 2700 c a l = - 2.7 c a l

Q. For the reaction : XX2​ OX4​(l)​2XOX2​(g) ΔU=2.1 k cal,ΔS=20 cal kX−1 at 300 K Hence ΔG is : -

2.7 k cal

- 2.7 k cal

9.3 k cal

- 9.3 k cal

Q. For the reaction :
$X X_{2} O X_{4} (l) 2 XO X_{2} (g)$
$ΔU = 2.1 k cal, ΔS = 20 cal k X^{- 1}$ at 300 K
Hence $Δ G$ is : -