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

A system at equilibrium contains I2(g) at a pressure of 0.46 atm and I(g) at a pressure of 0.34 atm . The system is then compressed to half its volume.

Part A: Find the pressure of I2 when the system returns to equilibrium.

Part B: Find the pressure of I when the system returns to equilibrium.

Thanks

2 Answers

Relevance
  • 2 years ago
    Favorite Answer

    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.

    • Login to reply the answers
  • 2 years ago

    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.)

    ====

    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.)

    • pisgahchemist
      Lv 7
      2 years agoReport

      "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" --- No. When the volume is cut in half, the pressure doubles. Pressure and volume are inversely proportional at constant temperature.

    • Login to reply the answers
Still have questions? Get your answers by asking now.