Question

In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction. N2(g) + 3H2(g) == 2NH3(g)

Answer 1

The equilibrium constant, Kc, for the given reaction can be calculated using the equilibrium concentrations of the reactants and products. The concentrations of nitrogen and hydrogen at equilibrium are 3.0 moles and 9.0 moles, respectively, while the concentration of ammonia is 18.0 moles. Using these values, the calculated equilibrium constant, Kc, is 0.1485.

Step-by-step explanation:

The equilibrium constant, Kc, for a chemical reaction is determined using the concentrations of the reactants and products at equilibrium. In this case, the reaction is N2(g) + 3H2(g) -> 2NH3(g). The given information states that two thirds of the nitrogen and hydrogen were converted into ammonia, which means the equilibrium concentrations of nitrogen and hydrogen can be calculated as 9.0 moles * (1 - 2/3) = 3.0 moles and 27.0 moles * (1 - 2/3) = 9.0 moles respectively. The equilibrium concentration of ammonia can be calculated as 27.0 moles * (2/3) = 18.0 moles. With these values, we can calculate Kc using the formula:

Kc = [NH3]^2 / ([N2] * [H2]^3)

Let's substitute the calculated concentrations into this formula to find Kc:

Kc = (18.0)^2 / (3.0 * 9.0^3)

Kc = 324 / 2187

Kc = 0.1485

Answer 2

The Kc for the reaction that we have here is 102

The equilibrium constant of a chemical reaction is a numerical value that expresses the ratio of the concentrations of products to reactants at a state of chemical equilibrium.

The value of is temperature-dependent, and it remains constant at a given temperature for a particular reaction.

Moles of nitrogen = 9.0 moles/ 10
`dm^3`

= 0.9 M

Moles of Hydrogen = 27 moles// 10
`dm^3`

= 2.7 M

Moles of ammonia = 2/3 * (0.9 + 2.7)

= 2.4 M

Kc =
`[NH_3]^2`/[
`[N_2] [H_2]^3`

Kc =
`[2.4]^2/[2.7]^3 [0.9]`

Kc = 5.76/17.7

Kc = 102