Question

Consider the reaction below. If the nitrosyl bromide (NOBr) is 17% dissociated at 55 degrees Celsius and the total pressure is 0.96 atm, what is K_p and K_c for the reaction at this temperature

2NOBr_(g)<-->2NO_(g)+Br_2(g)

Please show the work.

Answer 1

2NOBr(g) <-----> 2 NO(g) + Br2(g)

Initial x 0 0

17% dissociated.

at equilibrium x - 0.17x = 0.83x 2x 0.17 =0.34 x 2x0.17x = 0.34x

Hence,

total pressure at equilibrium = 0.83x + 0.34 x + 0.34 x = 1.51 x

But given that total pressure = 0.96 atm

Hence,

1.51 x = 0.96

x = 0.635 atm

Therefore,

Equilibrium partial pressures are

NOBr = 0.83x = 0.83 x  0.635 = 0.53 atm

NO = 0.34 x = 0.34 x 0.635 = 0.22 atm

Br2 = 0.34 x 0.635 = 0.22 atm

Kp = (0.22)² (0.22) / (0.53)²

= 0.038

Kp = 0.038

Kc:

Kp = Kc (RT)ⁿg

Kc= Kp/(RT)ⁿg

= 0.038 / (0.0821x 328K ) ^(2+1-2)

= 0.0014

Therefore,

Kc = 0.0014