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.

Explanation:

The balanced equilibrium reaction is

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

The expression for equilibrium reaction will be,

$K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}$

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

$K_p=\frac{(4.85)\times [(4.11)^4}{(5.04)^2\times (5.16)}$

$K_p=10.6$

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

$\Delta G^o=-2.303\times RT\times \log K_c$

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

$\Delta G^o=-2.303\times 8.314\times 298\times \log (10.6)$

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

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

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