Question

3/ At 2000 K , nitrogen gas and oxygen gas combine to form nitric oxide according to the following reaction: N2(g) + O2(g) ⇌ 2NO(g) ΔH = +1.81 KJ

with an equilibrium constant K of 4.1 x 10^-4. If 0.50 mole of N2 and 0.86 mole of O2 are put into a 2.0 L container at 2000 K , what would the equilibrium concentrations of all species be ?

4/ For the reaction in question #3. How would the following changes effect the equilibrium ? Each change occurs independently of the others. Increase temperature. Decrease pressure. Decrease [O2(g)]. Increase [NO(g)]

Answer 1

3)                                               N2(g)         +           O2(g)           ⇌        2NO(g)

     Inital concentration           0.5 / 2                       0.86 /2                             0

Change                                   -x                              -x                                   2x

Equilbrium concent              0.25-x                     0.43-x                             2x

Kc = (2x)² / (0.25 -x) ( 0.43-x) = 4.1 X 10^-4

4.1 X 10^-4 = 4x2 / (0.1075 -0.25x - 0.43 x + x²)

we can ignore x2 in denominator as it is very less as Kc is very low.

0.00041 (0.1075 - 0.68x ) = 4x2

0.0000441 - 0.000278x = 4x2

4x2 + 0.000278x -0.0000441 = 0

on solving

x = 0.00328

So equilibrium concentration of each species will be

[N2] = 0.25 - 0.00328 = 0.246

[O2] = 0.43-0.00328 =0.426

[NO] = 2X0.00328 = 0.00656

4) a) the enthalpy of reaction is positive so reaction is endothermic in nature, on increasing temperature the reaction will go forward direction ( Le-Chatelier's priciple)

b) decrease pressure will not affect the equilibrium as number of moles of reactants = number of moles of products = 2

c) decreasing conncentration of O2 ( or N2) will shift the equilbrium toward backward direction

d) Increase in concentration of NO (product) will shift the equilbrium toward backward direction