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

Answer:

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; $P_{[A] = 5x10^{-6}$, $P_{[B] = 0,99999atm$

Explanation:

For the reaction:

A(g) ⇄ 2B(g)

kp is: $kp = \frac{P_{[B]}^2}{P_{[A]}}$

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:

$1,4 = \frac{(1-2X)^2}{X}$

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:

$2,0x10^{-4} = \frac{(1-2X)^2}{X}$

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:

$2,0x10^5 = \frac{(1-2X)^2}{X}$

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!