Question

You mix an equal number of moles of nitrogen gas and hydrogen gas in a rigid container such that the total pressure is 2.0 atm. The gases react at a constant temperature to form ammonia and the system reaches equilibrium according to the equation:

N2(g) +3H2(g) = 2NH3(g)

a. At equilibrium, the total pressure is 1.7335 at a given temperature. Determine the value of Kp for this reaction at this temperature.

Answer 1

when we add equal number of mole
then mole fraction of each will be 0.5
P(H2) = (mole fraction)*(total pressure)
= 0.5*2
= 1 atm
P(N2) = (mole fraction)*(total pressure)
= 0.5*2
= 1 atm

reaction is
N2 + 3H2 < -- > 2NH3
1 1 0 intially
1-x 1-3x 2x at equlibrium

total pressure at equlibrium = 1-x+1-3x+2x
1.7335 = 2-2x
2x = 0.2665
x = 0.13325
= 0.133

so, at equlibrium
P(N2) = 1-X
= 1-0.13325
= 0.867 atm
P(H2) = 1-3X
= 1-3*0.133
= 0.601 atm
P(NH3) = 2x
= 2*0.133
= 0.266 atm

Kp = P(NH3)^2/(P(N2)*P(H2)^3)
= 0.266^2/(0.867*0.601^3)
= 0.377 atm^-2

Answer : 0.377 atm^-2