Question

28. 10. 10 mol of N,Q, is added to IL flask, what will be the concentrations atequilibrium? (quadratie equation required) N20, 2NO Initial Change Equilibrium 29. Find concentrations at equilibrium HA - HA 0.20 0 0 K. - 16 x 10 Initial Concentrations Change Equilibrium

Answer 1

28)

number of moles of N2O4 = 0.10 mol

volume = 1L

Concentration of N2O4 = number of moles/volume = 0.10/1.0 = 0.10M

                               N2O4 $\rightleftharpoons$ 2 NO2           K = 0.211

Initial                      0.10            0

change                    -x              +2x

equilibirum            0.10-x         +2x

K = [NO2]^2 / [N2O4]

0.211 = (2x)^2 / ( 0.10-x)

0.211 = 4x^2 / 0.10-x

0.0211- 0.211x = 4x^2

4x^2 + 0.211x - 0.0211=0

for solving the equation

x= 0.051

at equilibrium

concentration of N2O4 = 0.10-x = 0.10 - 0.051 = 0.049M

concentration of NO2 = 2x = 2x 0.051 = 0.102M

29)

                                     HA $\rightleftharpoons$ H+ + A-                Kc = 1.6 x10^-7

Initial                           0.20        0       0

change                       -x           +x      +x

equilibrium               0.20-x       +x      +x

Kc = [H+] [ A-] / [HA]

1.6 x10^-7 = x*x/(0.20-x)

for solving the equation

x= 0.000179

at equilibirum

[HA] = 0.20 - 0.000179 = 0.1998M

[H+] = 0.000179M = 1.79x10^-4M

[A-] = 0.000179M = 1.79x10^-4M