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 = 4.47 atm, P B = 5.17 atm, P C = 4.08 atm, and P D = 4.83 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘ C ?

The standard change in Gibbs free energy of this reaction at 25 ∘ C = 8.77 KJ/mol

Explanation:

                         3 A ( g ) + 4 B ( g ) ⇄ 2 C ( g ) + 3 D ( g )  -------(1)

Given

partial pressures (P A) = 4.47 atm, P B = 5.17 atm, P C = 4.08 atm, and P D = 4.83 atm.

                        From the above equation we can write  

                 ⇒  

                 ⇒  

                 ⇒

From equation (1)

              standard change in Gibbs free energy (ΔG°) = -RTlnKp

                                                                        = - 0.0821 x 298 x ln(0.029)

                                                                        = + 86.6 lit. atm / mole

∵ 1 lit.atm = 101.3 joule/mole

So, + 86.6 lit. atm  or 86.6 x 101.3 = 8772.58 joule/mole = 8.77 KJ/mol

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.

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

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