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 ?

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

Explanation:

For the given chemical equation:

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

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

where,

= standard Gibbs free energy = ?

R = Gas constant = 8.314 J/K mol

T = temperature =

= equilibrium constant at 25°C = 0.0632

Putting values in above equation, we get:

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

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