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Homework Help: Determine the Kc and mass of product obtained

  1. Dec 6, 2012 #1
    determine the Kc and mass of product obtained!!

    1. The problem statement, all variables and given/known data

    Consider a chemical reaction CO (g) + 2H2 (g) <-> CH3OH(g). When 2 moles of CO were reacting with 4 moles of H2 at temperature 500K and pressure p = 100bar, and chemical equilibrium was reached, n = 366.0 moles of CO was consumed.

    Based on the given data determine the value of the chemical equilibrium constant K(T) for T = 500K and find the mass (m) of the product obtained in the reaction.

    2. Relevant equations

    Kp = Kc (RT)Δn
    PV = nRT

    aA + bB <=> cC + dD

    Kp = [PC]c[PD]d / [PA]a[PB]b
    Kc = [C]c[D]d /[A]ab

    3. The attempt at a solution

    Okay so here is what I did. First the ice chart

    CO (g) + 2H2 (g) <-> CH3OH
    Ini 2mol 4mol -
    Ch -(.366) -2(.366) +(.366)
    Eq. 1.364 3.268 0.366 ∴Total Moles = 4.998 mol

    Eq. mol fraction = (Eq. mol / Total Mol)
    ∴ CO = 0.273 & H2 = 0.654 & CH3OH = 0.073
    Eq. partial pressure = Eq. mol frac X pressure
    ∴ CO = 26.9 atm & H2 = 64.5 atm & CH3OH = 7.2 atm

    Kp = [7.2]/([26.9][64.5]2) => 6.43 E -5
    ∴ Kc = Kp/(RT)Δn = 0.108

    To calculate mass I treated it as ideal gas so PV = nRT
    Where n = total mol = 4.998mol; T = 500K; P = 100 bar = 98.6 atm; R = 0.082 atm dm3 K-1 mol-1
    Total Volume = 2.08 dm3 = 2.08 L

    Now using density we can find the mass where density is 791.80g/L and d = m/V
    ∴ m = d x V = 791.80 x 2.08 = 1646.944 g

    Please let know if this problem was approached in the correct way or not.
     
  2. jcsd
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