Question

At 25 ∘C , the equilibrium partial pressures for the reaction were found to be PA=5.16 bar, PB=5.04 bar, PC=4.11 bar, and PD=4.85 bar . A(g)+2B(g)↽−−⇀4C(g)+D(g) What is the standard change in Gibbs free energy of this reaction at 25 ∘C ? Δ????∘rxn= kJmol

Answer 1

Answer: 5.85kJ/Kmol.

Step-by-step explanation:

The balanced equilibrium reaction is

`A(g)+2B(g)\rightleftharpoons 4C(g)+D(g)`

The expression for equilibrium reaction will be,

`K_p=\frac{[p_(D)]* [p_(C)]}^4{[p_(B)]^2* [p_(A)]}`

Now put all the given values in this expression, we get the concentration of methane.

`K_p=((4.85)* [(4.11)^4)/((5.04)^2* (5.16))`

`K_p=10.6`

Relation of standard change in Gibbs free energy and equilibrium constant is given by:

`\Delta G^o=-2.303* RT* \log K_c`

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature =
`25^0C=(25+273)K=298 K`

`K_c` = equilibrium constant = 10.6

`\Delta G^o=-2.303* 8.314* 298* \log (10.6)`

`\Delta G^o=5850.23J/Kmol`

`\Delta G^o=5.85kJ/Kmol`

Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.