Amixture initially contains a, b, and c in the following concentrations: [a] = 0.300 m , [b] = 1.05 m , and [c] = 0.550 m . the following reaction occurs and equilibrium is established: a+2b⇌c at equilibrium, [a] = 0.140 m and [c] = 0.710 m . calculate the value of the equilibrium constant, kc.

Kc = 9.52.

Explanation:

A + 2B ⇌ C,

Kc = [C]/[A][B]²,

Concentration:     [A]                [B]              [C]

At start:               0.3 M         1.05 M        0.55 M

At equilibrium:   0.3 - x        1.05 - 2x     0.55 + x

                            0.14 M        1.05 - 2x      0.71 M

∵ 0.3 M - x = 0.14 M.

∴ x = 0.3 M - 0.14 M = 0.16 M.

∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.

∵ Kc = [C]/[A][B]²

∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml