Question

2NOBr(g) ⇆ 2NO(g) + Br2(g) If nitrosyl bromide, NOBr, is 25.48 percent dissociated at 25°C and the total pressure is 0.325 atm, calculate KP and Kc for the dissociation at this temperature.

Answer 1

the partial pressure of the gas is directly proportional to the number of moles.

assume P be the initial pressure of the NOBr.

given equilibrium equation is

2 NOBr_(g) 2 NO(g) + Br_2(g)

ICE table for this reaction is

	NOBr	NO	Br2
Initial	P	0	0
Change	(-2x)	2x	x
equilibrium	(P-2x)	2x	x

since NOBr is 25.48% dissociated.

2x / P = 0.2548

P = 2x / 0.2548

P = 7.85 x

at equilibrium, the total pressure is 0.325 atm

P_NOBr + P_NO + P_Br2 = 0.325 atm

(P-2x) + 2x + x = 0.325

P - x = 0.325

7.85 x - x = 0.325

x( 7.85 - 1) = 0.325

6.85x = 0.325

x = 0.325 / 6.85 = 0.0474

the equilibrium partial pressures are

P_NOBr = P-2x = (7.85x - 2x) = (7.85 - 2)x = 5.85x = 5.85 0.0474 = 0.278 atm

P_NO =  2x = 2 0.0474 = 0.0948 atm

P_Br2 = x = 0.0474 atm

we can calculate Kp from these equilibrium pressures

given equilibrium equation is

2 NOBr_(g) 2 NO(g) + Br_2(g)

Kp = (P_NO)² (P_Br2) / (P_NOBr)²

Kp = (0.0948)² (0.0474) / (0.278)²

Kp = 0.00552 (or) 5.52 10^-3 atm

we know the relation between Kp and Kc

Kc = Kp / (RT)ⁿ

where, R = Ideal gas constant (0.082 atm.L/mol.K)

T = absolute temperature (K) = 298 K

n = Moles products - Moles reactants = 3 - 2 = 1

Kc = Kp / (RT)ⁿ

Kc = (0.00552) / (0.082 298)¹

Kc = 0.00552 / 24.436

Kc = 0.000225 (or) 2.25 10^-4 atm

so the answers are

Kp = 0.00552 (or) 5.52 10^-3 atm

Kc = 0.000225 (or) 2.25 10^-4 atm