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
