equilibrium

2NOBr(g) = 2NO(g) + Br2(g)
Initially:
No. of moles of NOBr = y mol
No. of moles of NO = No. of moles of Br2 = 0 mol

At eqm:
No. of moles of NOBr = y - (34%)y = 0.66y mol
No. of moles of NO = (34%)y = 0.34y mol
No. of moles of Br2 = (34%)y/2 = 0.17y mol
Total number of moles of all components = 0.66y + 0.34y + 0.17y = 1.17y
PNOBr = XNOBrPT = (0.66/1.17) x 0.25 = 0.25 x (66/117) atm
PNO = XNOPT = (0.34/1.17) x 0.25 = 0.25 x (34/117) atm
PBr2 = XBr2PT = (0.17/1.17) x 0.25 = 0.25 x (17/117) atm

Kp
= (PNO)2 (PBr2) / (PNOBr)2
= [0.25 x (34/117)]2 [0.25 x (17/117)] / [0.25 x (66/117)]2
= 0.00964 atm (or 974 Pa, as 1 atm = 1.01 x 105 Pa)

Concentration = n/V = P/RT
For NOBr:
PNOBr = 0.25 x (66/117) atm = 0.25 x (66/117) x (1.01x105) Pa
R = 8.314 J mol-1K-1
T = 273 + 25 = 298 K
[NOBr]eq = PNOBr/RT = = [0.25 x (66/117) x (1.01x105)] / (8.314 x 298) = 5.75 mol m-3
Similarly,
[NO]eq = PNO/RT = [0.25 x (34/117) x (1.01x105)] / (8.314 x 298) = 2.96 mol m-3
[Br2]eq = PBr2/RT = [0.25 x (17/117) x (1.01x105)] / (8.314 x 298) = 1.48 mol m-3

Hence. Kc
= [NO]eq2 x [Br2]eq / [NOBr]eq2
= (2.96)2 (1.48) / (5.75)2
= 0.392 mol m-3
= 0.000392 mol dm-3