Question

At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction. 2 NH3(g) N2(g) 3 H2(g) At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction?

Answer 1

Answer:

K = 3.37

Explanation:

2 NH₃(g) → N₂(g)  + 3H₂(g)

Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).

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

Initally       4moles             -            -

React        2moles           2m   +   3m

Eq             2 moles          2m        3m

We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)

The expression for K is:  ( [H₂]³ . [N₂] ) / [NH₃]²

We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)

K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²

K = 27/8 / 1 → 3.37

Answer 2

Answer:

The value of K for this reaction is 1.69

Explanation:

Step 1: Data given

Moles of NH3 = 4.0 moles

Volume of the container = 2.0 L

At the equilibrium 2.0 moles NH3 remains

Step 2: The balanced equation

2 NH3(g) → N2(g) + 3H2(g)

Step 3: Initial number of moles

NH3: 4.0 moles

N2: 0 moles

H2: 0 moles

Step 4: Number of moles at the equilibrium

NH3: 2.0 moles

This means there reacts 2.0 moles of NH3

For 2 moles of NH3 we have 1 mol of N2 and 3 moles of H2

There will be produced 1 mol of N2 and 3 moles of H2

Step 5: Calculate molarity

Molarity = moles / volume

Molarity of NH3 = 2.0 moles / 2.0 L = 1 M

Molarity of N2 = 1.0 mol / 2.0 L = 0.5 M

Molarity of H2 = 3.0 mol / 2.0 L = 1.5 M

Kc = ([H2]³[N2]) / [NH3]²

Kc = (1.5³ * 0.5) / (1²)

Kc = 1.69

The value of K for this reaction is 1.69