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

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.