Find the equilibrium partial pressures of a and b for each of the following different values of kp.? consider the following reaction: a(g) = 2b(g)find the equilibrium partial pressures of a and b for each of the following different values of kp. assume that the initial partial pressure of b in each case is 1.0 atm and that the initial partial pressure of a is 0.0 atm. make any appropriate simplifying assumptions. kp = 1.4? kp = 2.0 * 10^-4? kp = 2.0 * 10^5?

For Kp = 1,4; ,

For Kp = 2,0x10⁻⁴; ,

For Kp = 2,0x10⁵; ,

Explanation:

For the reaction:

A(g) ⇄ 2B(g)

kp is:

If initial pressure of B is 1,0atm and initial pressure of A is 0,0atm the equilibrium pressures are:

Replacing for Kp= 1,4:

1,4X = 4X² - 4X + 1

0 = 4X² - 5,4X + 1

Solving for X:

X = 0,22 -Right answer-

X = 1,13

Replacing:

For Kp= 2,0x10⁻⁴:

2,0x10^{-4}X = 4X² - 4X + 1

0 = 4X² - 4,0002X + 1

Solving for X:

X = 0,495atm

Replacing:

For Kp= 2,0x10⁵:

2,0x10^5X = 4X² - 4X + 1

0 = 4X² - 2,00004x10^5X + 1

Solving for X:

X = 5x10⁻⁶ -Right answer-

Replacing:

I hope it helps!