Question

At a certain temperature, 0.4011 mol of N 2 and 1.501 mol of H 2 are placed in a 3.00 L container. N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) At equilibrium, 0.1801 mol of N 2 is present. Calculate the equilibrium constant, K c . K c =

Answer 1

Answer:

Kc = 16.6

Explanation:

Step 1: Data given

Number of moles N2 = 0.4011 moles

Number of moles H2 = 1.501 moles

Volume = 3.00 L

At equilibrium, 0.1801 mol of N 2 is present

Step 2: The balanced equation

N2(g) + 3H2(g) ⇆ 2NH3(g)

Step 3: Calculate amount of N2 reacted

moles N2: 0.4011 mol - 0.1801 mol = 0.221 mol

For 1 mol N2 therewill react 3 moles of H2 to produce 2moles of NH3

Moles H2 reacted= 3 * 0.221 mol = 0.663 mol es

Moles NH3 produced = 2 * 0.221 mol = 0.442 mol es

Step 4: Amount of moles at equilibrium

Moles N2 = 0.1801 mol

[N2] = 0.1801 mol / 3.0 L = 0.06003 M

Moles H2 =1.501 moles - 0.663 mol = 0.838 moles

[H2] = 0.838 moles / 3.0 L = 0.2793 M

Moles NH3 = 0.442 mol

[NH3] = 0.442 mol / 3.0 L = 0.1473 M

Step 5: Calculate Kc

Finally, we can calculate Kc:

Kc = [NH3]² / ([N2] [H2]³)

Kc = (0.1473²)/(0.06003 *0.2793³)

Kc = 16.6