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Combustion and rate of reaction

  1. Nov 23, 2008 #1
    2 question I don't get ... =\ any help would be appreciated.
    I don't even know where to start ...

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

    2. A sample of a hydrocarbon is combusted completely in O2 to produce 21.83 g CO2, 4.47 g H2O and 311 kJ of heat. A) What is the mass of the hydrocarbon sample that was combusted? B) What is the empirical formula of the hydrocarbon? C) Calculate the value of ΔHfº per empirical-formula unit of the hydrocarbon.

    7. Suppose you have some reaction: 2A ---> B and B ---> 2A The rate for the formation of B from A is 3.8 x 102 s-1 and the rate for the formation of A from B is 8.6 x 10-1 s-1. A) What is the equilibrium constant for the equilibrium reaction, 2A <----> B ?
    B) Suppose that you started with 2 moles of A and at some later time you measure .9 moles of B in a total volume of 1L, is the reaction at equilibrium? Explain.

    2. Relevant equations

    3. The attempt at a solution

    2. mols C = 21.83/44 =.496

    mols H = 4.47/18.0 =,248 x2 =.496

    mols = g/Mwt then mols x Mwt = g

    The mass of C is .496 x12.0 = .5.97 g C
    The mass of H = .496 x 1.00 = ,496 g H
    Total mass is 6.47 g for the hydrocarbon

    The empirical formula is the ratio of the mols of C and H which makes it C1H1

    6.47/ 13 = .496 mols
    heat of combustion per formula weight is 311KJ/ .498 mols

    = 624.497992 kJ

    7. ???
    Last edited: Nov 23, 2008
  2. jcsd
  3. Nov 23, 2008 #2
    figured out number 2. Still don't get number 7
  4. Nov 24, 2008 #3


    User Avatar

    Staff: Mentor

    Have you heard about kinetic intepretation of equilibrium?

    Write expressions for
    - reaction equilibrium
    - forward reaction speed
    - backward reaction speed

    Compare denominator with forward reaction speed and denominator with backward reaction speed. Do you think you can combine rate coefficients to get equilibrium constnats?
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