Question

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?

Answer 1

For Kp = 1,4;
`P_{[A] = 0,22`,
`P_{[B] = 0,56atm`

For Kp = 2,0x10⁻⁴;
`P_{[A] = 0,495`,
`P_{[B] = 0,01atm`

For Kp = 2,0x10⁵;
`P_{[A] = 5x10^(-6)`,
`P_{[B] = 0,99999atm`

Step-by-step explanation:

For the reaction:

A(g) ⇄ 2B(g)

kp is:
`kp = (P_([B])^2)/(P_([A]))`

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

`P_{[A] = 0,0atm + X`

`P_{[B] = 1,0atm - 2X`

Replacing for Kp= 1,4:

`1,4 = ((1-2X)^2)/(X)`

1,4X = 4X² - 4X + 1

0 = 4X² - 5,4X + 1

Solving for X:

X = 0,22 -Right answer-

X = 1,13

Replacing:

`P_{[A] = 0,22`

`P_{[B] = 1,0atm - 0,44atm = 0,56atm`

For Kp= 2,0x10⁻⁴:

`2,0x10^(-4) = ((1-2X)^2)/(X)`

2,0x10^{-4}X = 4X² - 4X + 1

0 = 4X² - 4,0002X + 1

Solving for X:

X = 0,495atm

Replacing:

`P_{[A] = 0,495atm`

`P_{[B] = 1,0atm - 0,99atm = 0,01atm`

For Kp= 2,0x10⁵:

`2,0x10^5 = ((1-2X)^2)/(X)`

2,0x10^5X = 4X² - 4X + 1

0 = 4X² - 2,00004x10^5X + 1

Solving for X:

X = 5x10⁻⁶ -Right answer-

Replacing:

`P_{[A] = 5x10^(-6)`

`P_{[B] = 1,0atm - 0,00001atm = 0,99999atm`

I hope it helps!