Question

A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: NH_3(g) N_2(g) + H_2(g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K.

Answer 1

1.69M

Step-by-step explanation:

a balanced equation

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

looking at extent of dissociation of ammonia using the ice table we have NH3; 2M-2x. N2; -x, x H2; -3x, 3x

2M-2x=1

. 2x=1

. x=0.5

hence N2=0.5M and H2=3(0.5)=1.5M

k=([H2]^3[N2]^1)/[NH3]^2

k=1.69M

Answer 2

K = 1.69M

Step-by-step explanation:

For the reaction:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

K is defined as:

k = [N₂] [H₂]³ / [NH₃]² (1)

Molarity in equilibrium for each specie is:

NH₃(g): 2.0mol - 2x = 1.0mol/1L = 1M → X = 0.5mol

N₂(g): X = 0.5mol/1L = 0.5M

H₂(g): 3X = 1.5mol/1L = 1.5M

Replacing:

k = [0.5] [1.5]³ / [1]²

k = 1.69M

I hope it helps!