Question

At 800K, 2 mol of NO are mixed with 1 mol of O2. The reaction 2NO(g) + O2(g) <---> 2NO2(g) comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.71 mol of oxygen are present at equilibrium.

Calculate the equilibrium constant for the reaction.

answers 0.64

Answer 1

Given, the equilibrium reaction,

2NO(g) + O₂(g) 2NO₂(g)

Also given,

Initial moles of NO = 2 mol

Initial moles of O₂ = 1 mol

Moles of O₂ at equilibrium = 0.71 mol

Total pressure = 1 atm

Drawing an ICE chart,

	2NO(g)	O2(g)	2NO2(g)
I(moles)	2	1	0
C(moles)	-2x	-x	+2x
E(moles)	2-2x	1-x	2x

Now, Given,

Moles of O₂ at equilibrium = (1-x) = 0.71 mol

x = 0.29

Now, Calculating the moles at equilibrium for NO, NO₂,

Moles of NO at equilibrium = [ 2-2x] = 1.42 mol

Moles of NO₂ at equilibrium = [2x] = 0.58 mol

Now,

We know,

Mole fraction = Partial pressure / Total pressure

Thus, Calculating the mole fraction of each gas,

_NO = Moles of NO / Total moles

_NO = 1.42 / (1.42 + 0.58 + 0.71)

_NO = 0.524

Similarly,

_O2 = 0.71 / (1.42 + 0.58 + 0.71)

_O2 = 0.262

Also,

_NO2 = 0.58 / (1.42 + 0.58 + 0.71)

_NO2 = 0.214

Now, Calculating the partial pressures of each gas,

P_NO = _NOx Total pressure

P_NO = 0.524 x 1 atm

P_NO = 0.524 atm

Similarly,

P_O2 = 0.262 atm

P_NO2 = 0.214 atm

Now, the equilibrium constant expression is,

K_p = [P_NO2²] / [P_NO² x P_O2]

K_p = [0.214²] / [0.524² x 0.262]

K_p = 0.636

Thus, the equilibrium constant for the reaction is K_p = 0.64