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17 November, 11:26

The combustion products from burning pentane. CSH I2, with pure oxygen in a stone-. stoichiometric ratio exit at 2400 K, 100 kPa. Consider the dissociation of only CO2 and find the equilibrium mole fraction of CO.

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  1. 17 November, 11:53
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    The equilibrium mole fraction of CO is 0.0515

    Explanation:

    Solution

    Given that:

    Temperature T = 2400 K

    Pressure P = 100 kPa

    The reaction C₅H₁₂ + 8 O₂ →5CO₂ + 6 H₂ O

    Now,

    We take in K value from logarithm to the base e of the equilibrium constant K at 2400 K

    From reaction 2CO₂⇔ 2 CO + O₂

    In K = - 7.715

    P° = 0.1 MPa

    K = e^⁻7.715

    =4.461 * 10^⁻4

    Thus,

    2CO₂⇔ 2 CO + IO₂

    The initial mole CO₂ and shift reaction with 'x'

    The atom balance from the above reaction is given below:

    Species: CO₂ CO O₂ H₂O

    Initial: 5006

    Changes: - 2x 2xx0

    Total:5 - 2x 2xx6

    The total number of moles n tot = 5-2x+2x+x+6

    = 5+x+6

    =11+x

    Now,

    from the equation of the equilibrium constant K

    K = Y²co₂Yo₂/Y²co₂ (P/P₀)

    Yc₀ = nc₀/n tot

    = 2x/11+x

    Yco₂ = nc₀/n tot

    = 5-2x/11 + x

    So,

    K = (2x/1+x) ² * x/1+x / (5-2x/11 + x) ² * (100/100)

    4.461 * 10^⁻4 = (2x) ² / (11 + x) ² x / (11+x) (11+x) / (5-2x) * 1

    4.461 * 10^⁻4 = (2x/5-2x) ² x/11+x

    Now by trial and error method by keeping x = 0.291 the value satisfies the equation

    4.461 * 10^⁻4 = (2 * 0.291/5 - (2 * 0.291)) ² * (0.291/11 + 0.291)

    4.461 * 10^⁻4 = 4.47 * 10^⁻4

    Hence

    x = 0.291

    Thus,

    The mole fraction of CO₂ = nco₂/ntot

    Yco₂ = 5-2x-/11 + x

    Yco₂ = 5 - (2 * 0.291) / 11+0.291

    Yco₂ = 0.39128

    So,

    Mole fraction of CO = nco/ntot

    Yco = 2x/11+x

    =2 * 0.291/11+0.291

    Yco = 0.0515

    Then,

    Mole fraction of O₂ = no₂/ntot

    Yo₂ = x/11 + x

    = 0.291/11+0.291

    =0.02577
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