Question

At 25 ∘C25 ∘C,the equilibrium partial pressures for the reaction 3A(g)+2B(g)↽−−⇀C(g)+2D(g) 3A(g)+2B(g)↽−−⇀C(g)+2D(g) were found to be PA=5.84PA=5.84 atm, PB=4.47PB=4.47 atm, PC=4.17PC=4.17 atm, and PD=4.32PD=4.32 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘C25 ∘C ?

Answer 1

Answer:

9742.37 J/mol

Explanation:

We follow the expression for Gibbs free energy:

$\Delta G=\Delta G^{o}+RTlnQ$

First we calculate Q with the following expression for the reaction:

$aA+bB \rightleftharpoons cC+dD$

$Q=\frac{P_{C}^{c}P_{D}^{d}}{P_{A}^{a}P_{B}^{b}}$

Therefore, Q=0.0196

Now we continue with the first equation, and as we are on equilibrium, we know that the Gibbs free energy is zero, therefore:

$0=\Delta G^{o}+RTlnQ\\\\\Delta G^{o}=-RTlnQ\\\\R=8.314J/molK\\\\T=298 K\\\\\Delta G^{o}=+9742.37J/mol$