Chemistry problem (Mastering Chemistry)I have no idea how to do this one.?

Equilibrium pressure…..

It is important to note that this process occurs at a fixed temperature. This will keep the equilibrium constant the same.

2I(g) <==> I2(g) …………….. K = ?

0.34 atm…..0.46 atm ………. initial ….. K = 3.98

0.64 atm ….0.92 atm ………. when volume is cut in half, P’s momentarily jump to these values

-2x …………. +x ………………. changes to re-establish equilibrium

K = P(I2) / P²(I)

K = (0.46 atm) / (0.34 atm)² = 3.98

When the volume is cut in half, the pressures will double.

Q = (0.92 atm) / (0.68 atm)² = 1.99

Since Q is less than K, the system will shift to the right to re-establish equilibrium

K = P(I2) / P²(I)

3.98 = (0.92+x) / (0.68-2x)²

x = .088

P(I2) = 1.01 atm

P(I) = 0.504 atm

Check. The new partial pressures should give the value of K.

K = P(I2) / P²(I)

? = 1.01 atm / (0.504 atm)²

? = 3.98 …………………………… The new pressures check out.

Part A:

Consider the original equilibrium :

_______________ I₂(g) _____ ⇌ _____ 2I(g) ____ Kp

Original eqm: _ 0.46 atm _________ 0.34 atm

Kp = P(I)² / P(I₂)

Kp = 0.34² / 0.46 …… {1}

When the system is compressed to half its volume, the total pressure and both the partial pressures of I₂(g) and I(g) are reduced to half and thus the equilibrium is disturbed.

P(I₂) = (0.46 atm) × (1/2) = 0.23 atm

P(I) = (0.34 atm) × (1/2) = 0.17 atm

In the equation, there are 1 mole of gaseous molecules on the right, but 2 moles on the left.

According to Le Chatelier’s principle, when total pressure decreases, the equilibrium position shifts to the right in order to increase the number of gaseous molecules.

______________ I₂(g) _____ ⇌ _____ 2I(g) ____ Kp

Initial: _______0.23 atm ________ 0.17 atm

Change: ______ -y atm __________ +2y atm

New eqm: _(0.23 – y) atm ____ (0.17 + 2y) atm

When the new equilibrium is established:

Kp = (0.17 + 2y)² / (0.23 – y) …… {1}

As temperature is kept constant, Kp is unchanged, i.e. {2} = {1}

(0.17 + 2y)² / (0.23 – y) = 0.34² / 0.46

(4y² + 0.68y + 0.0289) / (0.23 – y) = 0.1156 / 0.46

0.46 × (4y² + 0.68y + 0.0289) = 0.1156 × (0.23 – y)

1.84y² + 0.3128y + 0.013294 = 0.026588 – 0.1156y

1.84y² + 0.4284y -0.013294 = 0

y = [-0.4284 ± √(0.4284 + 4×1..84×0.13294)] / (2×1.84)

y = 0.028 or y = -0.26 (rejected)

Pressure of I₂(g) at the new equilibrium = (0.23 – 0.028) atm = 0.202 atm ≈ 0.20 atm (to 2 sig. fig.)

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Part B:

Pressure of I(g) at the new equilibrium = (0.17 + 2×0.028) atm = 0.226 atm ≈ 0.23 atm (to 2 sig. fig.)