Question

For a reaction, X2(g) =2X(g), ∆U = 40 kJmol^-1, ∆S =
70 JK^-1mol^-1 at 600 K then ∆G will be

2.3 kJ mol^-1
1.95 kJ mol^-1
2.998 kJ mol^-1
-0.5 kJ mol^-1​

Answer 1

Answer:

The change in Gibbs Free energy is given by

ΔH=ΔU+Δn

g

RT

where

ΔH is the enthalpy of the reaction

ΔS is the entropy of the reaction

and ΔU is the change in internal energy

Δn

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,

ΔH=(2.1×10

3

)+(2×2×300)=3300cal

Hence,

ΔG=ΔH=TΔS

ΔG=3300−(300×20)

ΔG=−2700cal=−2.7cal