Calculating Equilibrium Concentrations 12 of 14 Carbonyl fluoride, COF2, is an i

Question

Calculating Equilibrium Concentrations 12 of 14 Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g) CO2(g)+CF4(g), K 9.00 The concentrations of reactants and products for a chemical reaction can be calculated if the equilibrium constant for the reaction and the starting concentrations of reactants and/or products are known. If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium? Express your answer with the appropriate units. View Available Hint(s) [COF2]ValueUnits ,- Submit Part B Consider the reaction CO(g) + NH3 (g):-HCONHe(g), Ke=0.870 If a reaction vessel initially contains only CO and NHs at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH2 be at equilibrium? Express your answer with the appropriate units. View Available Hint(s) HCONH2]Value Units

Explanation / Answer

                2CoF2(g) -------------> CO2(g) + CF4(g)
    I   2                    0        0

   C                  -2x                    +x       +x
          E              2-2x                       +x       +x
    
             Kc = [CO2][CF4]/[COF2]^2
            9    = x*x/(2-2x)^2

            9   = (x/2-2x)^2
             3   = x/2-2x

            3*(2-2x) = x

            6-6x-x = 0

           6-7x = 0

        x = 6/7 = 0.857
              x = 0.857
             [COF2]   = 2-2*0.857 = 2-2*0.857 = 0.286M
            [COF2]   = 0.286M >>>>answer

part-B
        CO(g) + NH3(g) ---------> HCONH2(g)
   I    1        2                  0
   C    -x       -x                 +x
   E    1-x     2-x                  +x

     Kc = [HCONH2]/[CO][NH3]
     0.87 = x/(1-x)(2-x)
     0.87*(1-x)(2-x) = x
        x = 0.556
  
    [HCONH2] =x = 0.556M

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