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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.