Question

Give detailed explanation please

Q2. Consider the reaction: 2NO (g)Cla(g) 2NOCIg) with Ke-6.6x10 at 38 C Calculate the equilibrium concentrations for all three species if starting concentration of NO, Cl2, and NOCI are 1.3 a) 10-3 M. 3.3 103 M, and 2.0 M, respectively. Show your work. b) Calculate Kp and the equilibrium partial pressure of all three species, when the starting partial pressure of NO, Cl2, and NOCl are 1.0 atm, 0.43 atm, and 1.75 atm, respectively. Show your work

Answer 1

a) 2NO + Cl2(g) <==> 2NOCl

Intial (M) 1.3×10-3 3.3×10-3. 2

Change. -2x. -x. +2x

Equilibrium 1.3x10^-3-2x 3.3×10^-3-x. 2+2x

Kc = [NOCl]²/[NO]²[Cl2]

6.6 × 10⁴ = (2+2x)²/((1.3x10^-3-2x)² × ( 3.3× 10^-3 -x))

Ignoring 4x³ , x = - 0.053 M ( acceptable) and 0.061 ( non acceptable, higher than 1.3 × 10-3 M). So reaction is going in backward direction. And equilibrium concentration are

NOCl = 2+ 2×( -0.053) = 1.892M

NO = 1.3 × 10-3 - 2×(- 0.053) = 0.1088M

Cl2 = 3.3 × 10-3 -(-0.053) = 0.0570 M

b) Kp = Kc(RT)^∆n

∆n = 2- 2-1 = 1, T = 273+38 = 311K, R = 8.314 J/Kmol

Kp = 6.6 × 10⁴ × (8.314 × 311) ^-1 = 25.52

Kp = [NOCl]²/[NO]²[Cl2]

(1.75+2x)²/(1-2x)²(0.43-x) = 25.52

x = 0.1399 = 0.14atm

NOCl = 1.75 + 2× 0.14 = 2.03 atm

Cl2 = 0.43- 0.14 = 0.29 atm

NO = 1- 2×0.14 = 0.72 atm