Question

In: Chemistry

The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g)



The dissociation of molecular iodine into iodine atoms is represented as

I2(g) ⇌ 2I(g)


At 1000 K, the equilibrium constant Kc for the reaction is 3.80

×

10−5. Suppose you start with 0.0461 mol of I2 in a 2.27−L flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2?

M

What is the equilibrium concentration of I?

M

Solutions

Expert Solution

                           I2(g) ⇌ 2I(g)

             I              0.0461         0

            C               -x               2x

            E            0.0461-x       2x

                [I-]   = 2x/2.27

                [I2]   = 0.0461-x/2.27

                  KC = [I-]^2/[I2]

                 3.8*10^-5 = (2x/2.27)^2/ 0.0461-x/2.27

                3.8*10^-5   = 4x^2/2.27(0.0461-x)

               3.8*10^-5 *(0.0461-x)*2.27 = 4x^2

                 x   = 0.000986

         [I-]   = 2x/2.27    = 2*0.000986/2.27   = 0.000868M

                [I2]   = 0.0461-x/2.27    = 0.0461-0.000986/2.27   = 0.0451/2.27 = 0.0198M

             

               


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