Question

For the reaction 2 NH 3 ( g ) − ⇀ ↽ − 3 H 2 ( g ) + N 2 ( g ) 2NH3(g)↽−−⇀ 3H2(g)+N2(g) the equilibrium concentrations were found to be [ NH 3 ] = 0.250 M [NH3]=0.250 M , [ H 2 ] = 0.700 M [H2]=0.700 M , and [ N 2 ] = 0.800 M [N2]=0.800 M . What is the equilibrium constant for this reaction?

Answer 1

Answer:

4.39

Explanation:

Let's consider the following reaction at equilibrium.

2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)

The concentration equilibrium constant (Kc) is the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc = [H₂]³.[N₂]/[NH₃]²

Kc = (0.700)³.(0.800)/(0.250)²

Kc = 4.39

Answer 2

Answer:

The equilibrium constant for this reaction is 4.39

Explanation:

Step 1: Data given

[NH3] =0.250 M

[H2] = 0.700 M

[N2] = 0.800 M

Step 2: The balanced equation

2NH3(g)⇆ 3H2(g) + N2(g)

Step 3: Calculate Kc

Kc = [N2][H2]³ / [NH3]²

Kc = (0.800 * (0.700³)) / (0.250²)

Kc = 4.39

The equilibrium constant for this reaction is 4.39