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29 November, 08:27

The equilibrium constant, Kp, for the following reaction is 9.52Ã10-2 at 350 K: CH4 (g) + CCl4 (g) 2CH2Cl2 (g). Calculate the equilibrium partial pressures of all species when CH4 and CCl4, each at an intitial partial pressure of 0.844 atm, are introduced into an evacuated vessel at 350 K. A) PCH4 = atm. B) PCCl4 = atm. C) PCH2Cl2 = atm.

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  1. 29 November, 09:18
    0
    A) p CH₄ = 0.732 atm

    B) p CCl₄ = 0.732 atm

    c) p CH₂Cl₂ = 0.22 atm

    Explanation:

    we have the equilibrium constant for this problem, along the initial pressures for the reactants, and need to find the partial pressures at equilibrium. So lets setup the equilibrium:

    CH₄ (g) + CCl₄ (g) ⇔ 2 CH₂Cl₂ (g) Kp = 9.50 x 10⁻²

    where Kp is given by:

    Kp = p CH₂Cl₂ ² / p CH₄ x p CCl₄

    where p are the partial pressures

    p CH₄ atm p CCl₄ atm p CH₂Cl₂ atm

    initial 0.844 0.844 0

    change - x - x + 2x

    equilibrium 0.844 - x 0.844 - x 2 x

    Kp = 9.52 x 10⁻² = (2x) ² / ((0.844 - x) x (0.844 - x))

    9.52 x 10⁻² = (2x) ² / (0.844 - x) ²

    Taking square root to both sides of the equation:

    √9.52 x 10⁻² = 2x / (0.844 - x)

    0.309 = 2x / (0.844 - x)

    0.260 - 0.309 x = 2x

    0.260 = 2.309 x ⇒ x = 0.112

    So the partial pressures are at equilibrium are:

    p CH₄ = p CCl₄ = 0.844 - 0.112 = 0.732 atm

    p CH₂Cl₂ = 2 x (0.112 atm) = 0.224 atm

    You can check your answer is correct by plugging this values and comparing with the given Kp.
  2. 29 November, 09:40
    0
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    impact and deflation

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    deposition and erosion

    accumulation and compaction
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