Question

Hey guys..!!!!
what is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICI was 0.78 M?​

Answer 1

2ICl (g) <=> I2(g) + Cl2(g) ; Kc = 0.14

at t = 0

conc. of ICl = 0.78 M

conc. of I2 = 0 M

conc. of Cl2 = O M

at equilibrium ,

conc. of ICl = (0.78 - 2x ) M

conc. of I2 = x M

conc. of Cl2 = x M

now, Kc = [I2][Cl2]/[ICl]²

Kc = x.x/(0.78 - 2x )²

0.14 = x²/(0.78 - 2x )²

take square root both sides,

√0.14 = x/(0.78 - 2x )

0.374 = x/(0.78 - 2x )

x = 0.29172 - 0.748x

1.748x = 0.21972

x = 0.29172/1.748 = 0.1668 ≈ 0.17

hence, conc. of ICl at equilibrium = 0.78 - 0.34 = 0.44 M

conc. of I2 = x = 0.17 M

conc. of Cl2 = x = 0.17 M

Answer 2

Answer:

2ICl (g) <=> I2(g) + Cl2(g) ; Kc = 0.14

at t = 0

conc. of ICl = 0.78 M

conc. of I2 = 0 M

conc. of Cl2 = O M

at equilibrium ,

conc. of ICl = (0.78 - 2x ) M

conc. of I2 = x M

conc. of Cl2 = x M

now, Kc = [I2][Cl2]/[ICl]²

Kc = x.x/(0.78 - 2x )²

0.14 = x²/(0.78 - 2x )²

take square root both sides,

√0.14 = x/(0.78 - 2x )

0.374 = x/(0.78 - 2x )

x = 0.29172 - 0.748x

1.748x = 0.21972

x = 0.29172/1.748 = 0.1668 ≈ 0.17

hence, conc. of ICl at equilibrium = 0.78 - 0.34 = 0.44 M

conc. of I2 = x = 0.17 M

conc. of Cl2 = x = 0.17 M