equilibrium?

Question

The following reaction has Kp = 109 atm-1 at 258°C.
2 NO(g) + Br2(g) 2 NOBr(g)

If the equilibrium partial pressure of Br2 is 0.0108 atm and the equilibrium partial pressure of NOBr is 0.0717 atm, calculate the partial pressure of NO at equilibrium.

Answer 1

If you have the reaction:

aA + bB ↔ cC + dD

Then:

Kp = ((PC)^c * (PD)^d) / ((PA)^a * (PB)^b) (PC is the partial pressure of C, PD is the partial pressure of D, etc.)

Kp = 109 = ((PNOBr)^2) / ((PNO)^2 * (PBr2)^1)

109 = (0.0717)^2 / ((PNO)^2 * 0.0108)

1.18(PNO)^2 = 0.00514

(PNO)^2 = 0.00436

PNO = 0.0660 atm

*** The partial pressure of NO at equilibrium = 6.60 x 10^(-2) atm