Question

Consider the following equilibrium: 2NOCl(g) ⇌ 2 NO(g) + Cl2(g)              Kc = 1.6 x 10-5 1.00...

Consider the following equilibrium:

2NOCl(g) ⇌ 2 NO(g) + Cl2(g)              Kc = 1.6 x 10-5

1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in a 1.00 L container. Calculate the equilibrium concentration of NO(g)

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Answer #1

concentration of NOCl [NOCl]   = no of moles/volume in L

                                                  = 1/1 = 1M

concentration of Cl2 [Cl2]   = no of moles/volume in L

                                                  = 1/1 = 1M

             2NOCl(g) ⇌ 2 NO(g) + Cl2(g)

I              1                   0               0

C            -2x                 2x             x

E            1-2x               2x              x

                 Kc     = [NO]^2[Cl2]/[NOCl]^2

                1.6*10^-5   = (2x)^2*x/(1-2x)^2

                1.6*10^-5 *(1-2x)^2 = 4x^3

                   x   = 0.01554

[NO]   = 2x    = 2*0.01554 = 0.03108M >>>>answer

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