Question

A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of NH3(g) into a 1.00 liter container, she found the equilibrium concentration of H2(g) to be 0.103 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =

Answer 1

The equilibrium constant (Kc) for the decomposition reaction of NH3 into N2 and H2 is calculated by first determining the changes in concentrations of reactants and products at equilibrium, and then substituting these values into the equilibrium constant expression.

Step-by-step explanation:

The question asks for the calculation of the equilibrium constant (Kc) for the decomposition of ammonia (NH3) into nitrogen (N2) and hydrogen (H2) gases at a specific temperature of 671 K. Given that 7.33 × 10-2 moles of NH3 were introduced into a 1.00-liter container and the equilibrium concentration of H2 was found to be 0.103 M, we can use the reaction stoichiometry to find the equilibrium concentrations of N2 and NH3, and then apply them to the equilibrium constant expression.

The balanced equation for the reaction is: 2 NH3(g) ⇌ N2(g) + 3 H2(g).

Lets assume that the change in concentration of NH3 is x, then the change in concentration of N2 is x/2 and for H2 it is 3x/2. Using the equilibrium concentration of H2, we can solve for x.

The equilibrium concentration of NH3 would be (0.0733 - 2x) and for N2 it would be x/2. Substituting these values into the expression for Kc, which is [N2] [H2]3 / [NH3]2, we can calculate the value of Kc.

Answer 2

Kc = 8.05x10⁻³

Step-by-step explanation:

This is the equilibrium:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Initially 0.0733

React 0.0733α α/2 3/2α

Eq 0.0733 - 0.0733α α/2 0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³