Question

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

Part A Kp= 2.0

Enter your answers numerically separated by a comma. Express your answer using two significant figures

Part B

Kp= 1.8×10⁻⁴

Enter your answers numerically separated by a comma. Express your answer using two significant figures.

Part C

Kp= 1.6×10⁵

Enter your answers numerically separated by a comma. Express your answer using two significant figures.

Answer 1

Given, the equilibrium reaction,

A(g) 2B(g)

The initial partial pressure of A = 0.0 atm

The initial partial pressure of B = 1.0 atm

Part A) Given,

K_p = 2.0

Now, reversing the given reaction,

2B(g) A(g)

Thus, K_p'= 1/ K_p

K_p'= 1/ 2.0

K_p'= 0.5

Now, Drawing an ICE chart,

	2B(g)	A(g)
I(atm)	1.00	0.0
C(atm)	-2x	+x
E(atm)	1.00-2x	x

Now, the K_p' expression is,

K_p'= P_A / P_B²

0.5 =  x / (1.00-2x)²

0.5 =  x / (1.00-4x + 4x²)

2x² -3x + 0.5 = 0

Solving the quadratic equation,

x = 0.191

Thus, from ICE chart,

P_A = x = 0.19 atm

P_B = (1.00-2x) = 0.62 atm

Part B) Given,

K_p = 1.8 x 10^-4

Now, reversing the given reaction,

2B(g) A(g)

Thus, K_p'= 1/ K_p

K_p'= 1/ (1.8 x 10^-4)

K_p'= 5.5 x 10³

Now, Drawing an ICE chart,

	2B(g)	A(g)
I(atm)	1.00	0.0
C(atm)	-2x	+x
E(atm)	1.00-2x	x

Now, the K_p' expression is,

K_p'= P_A / P_B²

5.5 x 10³ =  x / (1.00-2x)²

5.5 x 10³​​​​​​​ =  x / (1.00-4x + 4x²)

(2.22 x 10⁴)x² -(2.22 x 10⁴)x + (5.5 x 10³)​​​​​​​ = 0

Solving the quadratic equation,

x = 0.496

Thus, from ICE chart,

P_A = x = 0.50 atm

P_B = (1.00-2x) = 0.0 atm

Part C) Given,

K_p = 1.6 x 10⁵

Now, reversing the given reaction,

2B(g) A(g)

Thus, K_p'= 1/ K_p

K_p'= 1/ (1.6 x 10⁵)

K_p'= 6.25 x 10^-6

Now, Drawing an ICE chart,

	2B(g)	A(g)
I(atm)	1.00	0.0
C(atm)	-2x	+x
E(atm)	1.00-2x	x

Now, the K_p' expression is,

K_p'= P_A / P_B²

6.25 x 10^-6 =  x / (1.00-2x)²

6.25 x 10^-6​​​​​​​ =  x / (1.00-4x + 4x²)

(2.5 x 10^-5)x² - x + (6.25 x 10^-6)​​​​​​​ = 0

Solving the quadratic equation,

x = 6.25 x 10^-6

Thus, from ICE chart,

P_A = x = 6.2 x 10^-6 atm Or 6.3 x 10^-6 atm

P_B = (1.00-2x) = 1.0 atm