Verizon LTE 19:10 80% At equilibrium, the concentrations of reactants and produc

Question

Verizon LTE 19:10 80% At equilibrium, the concentrations of reactants and products can be predicted using the equilibrium constant, Ke, which is a mathematical expression based on the chemical equation. For example, in the reaction where a, b, c, and d are the stoichiometric coefficients, the equilibrium constant is where [A]·[B], [C]. and [D] are the eqibrum, concentrations, if the reactors not at equilibrium, the quantity can still be calculated, but it is called the reaction quotient,Qc nstead of the equilibrium constant, K where each concentration is measured at some arbitrary time t Part A A mixture initially contains A, B, and C in the folowing concentrations: A -0.300 M B-1.25 M, and C)-0.350 M.The following reaction occurs and equilibrium is established At equilibrium, A-0.130 M and C-0.520 M.Calculate the value of the equilibrium constant, K Express your answer numerically. View Available Hints) Provide Feedback When conducting chemical reactions in the lab or in industrial processes, it can be important t now whether a reaction has reached equilibrium. By measuring the reaction quotient, Q. of a hemical reaction and comparing it to the equilibrium constant, K, we can identify whether a eaction is at equilibrium. For the reaction he reaction quotient, Q. is given by the expression A B nder any conditions. The value of Q is equal to K only at eqbnn. Part A

Explanation / Answer

part-A
               A     + 2B ----------------> C
         I    0.3       1.25                 0.35
         C   -0.17     -2*0.17              +0.17
         E    0.13      0.91                 0.52
       
               KC = [c]/[A][B]^2
                   = 0.52/0.13*(0.91)^2    = 4.83
part-A
         N2(g)+ 3H2(g) -------> 2NH3(g)
        Qc = 3.56*10^-4
         KC = 6.02*10^-2
       KC>QC The reaction will procede to right side.
The reaction is not t equilibrium and will procede to right.
part-B
       
           C(s) + H2O(g) -------------> Co(g) + H2(g)
         
           C is solid. [C] =1
             [H2O] = no of moles/volume in L
                    = 13.6/3 = 4.53M
              [CO] = no of mole/volume in L
                    = 3.8/3   = 1.26M
               [H2] = no of moles/volume in L
                     = 7.9/3 = 2.63M

          Qc = [CO][H2]/[H2O]
              = 1.26*2.63/4.53   = 0.73

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