Question

Will literally worship you if you can give me some assistance with these. (I rate immediately)

Consider the reaction. A(g)⇌2B(g) Find the equilibrium partial pressures of A and B for each of the 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= 1.6

Part B

Kp= 1.5×10−4

Part C

Kp= 1.6×105

Answer 1

Part A)

Given, the equilibrium reaction,

A(g) 2B(g) -------K_p = 1.6

P_A = 0.0 atm

P_B = 1.0 atm

Reversing the reaction,

2B(g) A(g)

	2B(g)	A(g)
I(atm)	1.0	0
C(atm)	-2x	+x
E(atm)	1.0-2x	x

Now, K_p' = 1 /K_p

K_p' = 1 /1.6

K_p' = 0.625

Now, the K_p' expression is,

K_p' = P_A / P_B²

K_p' = x / (1-2x)²

0.625 = x / ( 1 -4x +4x²)

0.625 - 2.5x + 2.5x² = x

2.5x² - 3.5 x + 0.625 = 0

Solving the quadratic equation,

x = 0.2101

Thus,

P_A = x = 0.21 atm [ 2s.f]

P_B = ( 1-2x) = 0.58 atm [2 s.f]

Part B)

Given, the equilibrium reaction,

A(g) 2B(g) -------K_p = 1.5 x 10^-4

P_A = 0.0 atm

P_B = 1.0 atm

Reversing the reaction,

2B(g) A(g)

	2B(g)	A(g)
I(atm)	1.0	0
C(atm)	-2x	+x
E(atm)	1.0-2x	x

Now, K_p' = 1 /K_p

K_p' = 1 /(1.5 x 10^-4)

K_p' = 6666.667

Now, the K_p' expression is,

K_p' = P_A / P_B²

K_p' = x / (1-2x)²

6666.667 = x / ( 1 -4x +4x²)

6666.667 - 26666.67x + 26666.67x² = x

26666.67x² - 26667.67x + 6666.667 = 0

Solving the quadratic equation,

x = 0.50

Thus,

P_A = x = 0.50 atm [ 2s.f]

P_B = ( 1-2x) = 0.0 atm

Part C)

Given, the equilibrium reaction,

A(g) 2B(g) -------K_p = 1.6 x 10⁵

P_A = 0.0 atm

P_B = 1.0 atm

Reversing the reaction,

2B(g) A(g)

	2B(g)	A(g)
I(atm)	1.0	0
C(atm)	-2x	+x
E(atm)	1.0-2x	x

Now, K_p' = 1 /1.6 x 10⁵

K_p' = 1 /(1.5 x 10^-4)

K_p' = 6.25 x 10^-6

Now, the K_p' expression is,

K_p' = P_A / P_B²

6.25 x 10^-6 = x / (1-2x)²

6.25 x 10^-6 = x / (1)² ------Here (1-2x) 1 Since, x<<< 1

x = 6.25 x 10^-6

Thus,

P_A = x = 6.2 x 10^-6 atm Or 6.3 x 10^-6 atm [ 2s.f]

P_B = ( 1-2x) = 1.0 atm Or 0.99 atm