Question

At 25 ∘ C , the equilibrium partial pressures for the reaction 3 A ( g ) + 4 B ( g ) − ⇀ ↽ − 2 C ( g ) + 3 D ( g ) were found to be P A = 5.70 atm, P B = 4.00 atm, P C = 4.22 atm, and P D = 5.52 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘ C ?

Answer 1

Answer: The standard Gibbs free energy of the given reaction is 6.84 kJ

Explanation:

For the given chemical equation:

$3A(g)+4B(g)\rightleftharpoons 2C(g)+3D(g)$

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_C)^2\times (p_D)^3}{(p_A)^3\times (p_B)^4}$

We are given:

$(p_A)_{eq}=5.70atm\\(p_B)_{eq}=4.00atm\\(p_C)_{eq}=4.22atm\\(p_D)_{eq}=5.52atm$

Putting values in above expression, we get:

$K_p=\frac{(4.22)^2\times (5.52)^3}{(5.70)^3\times (4.00)^4}\\\\K_p=0.0632$

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = 8.314 J/K mol

T = temperature = $25^oC=[273+25]K=298K$

$K_{eq}$ = equilibrium constant at 25°C = 0.0632

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.0632)\\\\\Delta G^o=6841.7J=6.84kJ$

Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ