The reaction
2PH3(g)+As2(g)⇌2AsH3(g)+P2(g)
has Kp=2.9×10−5
at 873 K. At the same temperature, what is Kp for each of the following reactions?

Parts A, B, and C

Part A:
2AsH3(g)+P2(g)⇌2PH3(g)+As2(g)

Part B:

6PH3(g)+3As2(g)⇌3P2(g)+6AsH3(g)

Part C:

2P2(g)+4AsH3(g)⇌2As2(g)+4PH3(g)

Part A K´p = 3.4 x 10⁴

Part B K´p = 2.4 x 10⁻¹⁴

Part C K´p = 1.2 x 10⁹

Explanation:

The methodolgy to answer this question is to realize that the equilibrium in part A is the reverse of the the equilibrium given:

2PH₃(g) + As₂(g)  ⇌ 2 AsH₃(g) + P₂(g)  Kp = 2.9 x 10⁻⁵

Kp = [AsH₃]²[P₂]/[PH₃]²[As]  = 2.9 x 10⁻⁵

Likewise, part B is this equilibrium where the coefficients have been multiplied by 3,  and part C is  is the  reverse equilibrium multipled by 2.

Given the way that the equilibrium constant, Kp, is expressed mathematically, we can see that if we reserse the equilibrium its constant will be the inverse of the old Kp.

Similarly if we multiply the coefficient by a factor we have to raise the expression for Kp to that factor.

With that in mind lets answer the three parts of the questiion.

Part A

Lets call the new equilibrium K´p :

K´p = 1/ Kp = 1/2.9 x 10⁻⁵ = 3.4 x 10⁴

Part B

K´p = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴

Part C

K´p = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹

If you can not see why this is so, lets show it for part C

K´p = [As₂]²[PH₃]⁴ / [P₂]²[AsH₃]⁴

rearranging this equation:

K´p = (1 / {  [AsH₃]⁴[P]² / [As₂]²[PH₃]⁴ } =) 1 / Kp²

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.