Question

If you put 0.150 M NO2 and 0.400 M H2 in a reaction vessel, you end up with 0.0750 M NH3 at equilibrium:

2 NO2(g) + 7 H2(g) ↔ 2 NH3(g) + 4 H2O(g)

1.What is the value of the equilibrium constant, Kc?

2.What is the NH3 concentration at equilibrium?

Answer 1

Solution :-

Initial concentration 0.150 M NO2 and 0.400 M H2

Equilibrium concentration of NH3 = 0.0750 M

Using the ICE table we can find the value of x

2NO2   + 7H2   ----- > 2 NH3 + 4 H2O

0.150 M    0.400 M         0               0

-2x                -7x               +2x           +4x

0.150-2x    0.400-7x       0.0750        4x

Using the equilibrium value of NH3 we can find the value of x

2x=0.0750 M

X=0.0750 M /2 = 0.0375 M

Now using the value of x we can find the equilibrium concentration of each species

[NO2] eq = 0.150-2x = 0.150 – (2*0.0375) = 0.075 M

[H2]eq = 0.400-7x = 0.400- (7*0.0375) = 0.1375 M

[NH3] = 2x = 2*0.0375 = 0.075 M

[H2O] = 4x = 4*0.0375 = 0.150 M

Now lets calculate the equilibrium constant Kc

Kc=[NH3]^2[H2O]^4/ [NO2]^2[H2]^7

Kc=[0.075]^2 [0.150]^4 / [0.075]^2 [0.1375]^7

Kc = 545

Therefore the equilibrium constant Kc for the reaction is 545

Part 2) The equilibrium concentration of the NH3 is 0.0750 M