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 CPart A:2AsH3(g)+P2(g)⇌2PH3(g)+As2(g)Part B:6PH3(g)+3As2(g)⇌3P2(g)+6AsH3(g)Pa rt C:2P2(g)+4AsH3(g)⇌2As2(g)+4PH3(g)

Part A

Kp = 3.4 x 10⁴

Part B

Kp = 2.4 x 10⁻¹⁴

Part C

Kp = 1.2 x 10⁹

Explanation:

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

Part A

it is the inverse of the equilibrium given

Kp(A) = 1/ Kp = 1 / 2.9 x 10⁻⁵ = 3.4 x 10⁴

Part B

Is the equilibrium where the coefficients have been multiplied by 3,

Kp(B) = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴

Part C

This is the  reverse equilibrium multipled by 2.

Kp(C) = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹

kudos to this man hope he is right if not sorry for taking the slot

Explanation: