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Math Help - chemistry questions

  1. #1
    Senior Member euclid2's Avatar
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    chemistry questions

    1. what is the value of the equilibrium constant at 200^oC for the decomposition of phosporus pentachloride gas to phosphorus trichloride gas and chlorine gas? At equilibrium,

    [PCL_{3(g)}]=[CL_{2(g)}]=0.014 mol/L

    [PCL_{5(g)}] = 4.3*10^{-4} mol/L

    4. In the haber process of synthesizing ammonia, the value of K is 8.00 * 10^{-7} for the reaction at 400^oC. In a sealed container at equilibrium at 400^oC the concentations of H_{2(g)} and N_{2(g)} are measured to be 0.50 mol/L and 1.50mol/L respectively. Write the equilibrium constant equation and calculate the equilibreum concentration of NH_{3(g})
    Last edited by euclid2; December 31st 2008 at 10:03 PM.
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  2. #2
    Super Member fardeen_gen's Avatar
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    1. what is the value of the equilibrium constant at for the decomposition of phosphate pentachloride gas to phosphorus trichloride and chlorine gas? At equilibrium,

    The reaction is:
    PCl5 <---> PCl3 + Cl2
    At equilibrium: PCl5=(4.3 x 10^-4) PCl3 =0.014 Cl2= 0.014

    Therefore,
    Equilibrium constant K
    = 0.014 x 0.014/(4.3 x 10^-4)
    = 0.000196/(4.3 x 10^-4)
    =0.4558
    Last edited by fardeen_gen; December 31st 2008 at 10:35 PM. Reason: Incorporated new info. regarding the question introduced by the poster later
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  3. #3
    Senior Member euclid2's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    The reaction is:
    PCl5 ---> PCl3 + Cl2
    Initial conc. : 1 0 0
    At equilibrium: 1 - x x x

    Therefore, according to given data, x = 0.014 mol/L

    Therefore,
    Equilibrium constant K
    = x.x/(1 - x)
    = x^2/(1 - x)
    = (0.014)^2/(1 - 0.014)
    = 0.000196/0.086
    = 0.0023
    You were missing a bit of info, i edited, sorry.
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  4. #4
    Super Member fardeen_gen's Avatar
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    4. In the haber process of synthesizing ammonia, the value of K is for the reaction at . In a sealed container at equilibrium at the concentations of and are measured to be 0.50 mol/L and 1.50mol/L respectively. Write the equilibrium constant equation and calculate the equilibreum concentration of )
    N2 + 3H2 <---> 2NH3
    Equilibrium concentration: [H2] = 0.50 M, [N2] = 1.50 M
    Let the equilibrium concentration of NH3 be x M

    Therefore,
    Equilibrium constant equation: (By law of mass action)
    K = [NH3]^2/([H2]^3 x [N2])
    8 x 10^-7 = x^2/((0.5)^3 x (1.5))
    x^2 = (8 x 10^-7) x (0.125) x (1.5)
    x^2 = 1.5 x 10^-7
    x = 3.87 x 10^-4
    Last edited by fardeen_gen; December 31st 2008 at 10:35 PM. Reason: Its a reversible reaction!I forgot to show a double sided arrow!
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  5. #5
    Senior Member euclid2's Avatar
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    Equilibrium constant equation: (By law of mass action)
    K = [NH3]^2/([H2]^3 x [N2])
    x^2 = (8 x 10^-7) x (0.125) x (1.5)
    these
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  6. #6
    Super Member fardeen_gen's Avatar
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    Can you quote the exact lines please?

    EDIT:
    By the Gulbarg-Waage law of mass action:
    For a reaction:

    where, [S], [T], [A], [B] are equlibrium concentrations and α,β, σ, τ are molar coefficients(number of moles of the particular element or compound, for example, σ moles of S are formed in the above equation)



    Now, since
    the value of K is 8 x 10^-7 for the reaction at 400 degree celsius
    , therefore as the required reaction also takes place at 400 degree celsius, we can use the above value of K and plug it into the equilibrium constant equation. Then all I have done is some cross multiplication(using the given values) to get concentration of NH3(which I assumed to be x moles/L)
    Last edited by fardeen_gen; December 31st 2008 at 10:34 PM.
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  7. #7
    Senior Member euclid2's Avatar
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    Equilibrium constant equation: (By law of mass action)
    K = [NH3]^2/([H2]^3 x [N2])
    8 x 10^-7 = x^2/((0.5)^3 x (1.5))
    x^2 = (8 x 10^-7) x (0.125) x (1.5)
    these
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  8. #8
    Senior Member euclid2's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Can you quote the exact lines please?

    EDIT:
    By the Gulbarg-Waage law of mass action:
    For a reaction:

    where, [S], [T], [A], [b] are equlibrium concentrations and α,β, σ, τ are molar coefficients(number of moles of the particular element or compound, for example, σ moles of S are formed in the above equation)

    thanks for your time, and happy new year!
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