Question

Nitric Oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. The reaction is

2NO(g)+Br₂(g)<->2NOBr(g)

What is the value of K_p? What would the partial pressures be if NO and Br_2, both at an initial partial pressure of .30 atm, were allowed to come to equilibrium at this temperature?

Answer 1

For the reaction 2NO + Br2----> 2NOBr

intial pressure of NO= 98.4 torr and Br2=41.3 Torr

let drop in pressure of Br2 due to equilibrium =x

So at equilibrium NO= 98.4-2x and NOBr= 2x and Br2=41.3-x

total pressure = 98.4-2x+2x+41.3-x =139.7-x= 110.5

x=139.7-110.5=29,2 torr

Partial pressures at Equilibrium :NO : 98.4-2*29.2=40 Torr and NOBr= 2*29.2= 58.4 torr and Br2= 41.3-29.2= 12.1 torr

Kp = (58.4)²/ {40*40* 12.1)= 0.1761

b) let P= drop in partial pressure

therefore NO= 0.3*760 torr-2P =228-2P and Br=0.3*760-P =228-P and NOBr= 2P

Kp =4P²/{ (228-P)*(228-2P) =0.1761

soving this gives P= 37.7 torr

Partial pressures of NO= 0.3*760-2*37.7=152.6 Torr and Br2= 228-37.7=200.3 torr