A student ran the following reaction in the laboratory at 30

A student ran the following reaction in the laboratory at 302 K: 2NO(g) + Br2(g) 2NOBr(g) When she introduced NO(g) and Br2(g) into a 1.00 L evacuated container, so that the initial partial pressure of NO was 1.34 atm and the initial partial pressure of Br2 was 0.508 atm, she found that the equilibrium partial pressure of NOBr was 0.667 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction.

Solution

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

Initial 1.34 0.508 0

at equilibrium 1.34-2X 0.508-X 2X

given pNOBr equilibrium = 0.667 = 2X , hence X= 0.3335 ,

pNO = 1.34 - 0.667 = 0.673 , pBr2 = 0.1745

Kp = ( pNOBr)^2 / ( pNO^2 x pBr2)

= ( 0.667^2) / ( 0.673^2 x 0.1745)

= 5.63