At a certain temperature, 0.4611 mol of N 2 and 1.661 mol of H 2 are placed in a 4.50 L container. N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽

Question

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At a certain temperature, 0.4611 mol of N 2 and 1.661 mol of H 2 are placed in a 4.50 L container. N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽

− 2 NH 3 ( g ) N 2 ( g ) + 3 H 2 ( g ) ↽ − − ⇀ 2 NH 3 ( g ) At equilibrium, 0.1801 0.1801 mol of N 2 N 2 is present. Calculate the equilibrium constant, K c K c .

Chemistry

1 answer:

Monica [59] · Answer 1 · 2022-02-26T20:49:20+03:00

Answer:

The equilibrium constant is 64.76

Explanation:

Step 1: Data given

Number of moles of N2 = 0.4611 moles

Number of moles of H2 = 1.661 moles

Volume = 4.50 L

At the equilibrium there is 0.1801 mol of N2

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3 (g)

Step 3: The initial moles

N2: 0.4611 mol

H2: 1.661 mol

NH3: 0 mol

Step 4: Calculate the moles reacted

N2: 0.4611 mol - 0.1801 mol = 0.281 mol

H2: 3 * 0.281 mol = 0.843 mol

NH3: 2 * 0.281 mol = 0.562 mol

Step 5: Calculate moles at the equilibrium

N2: 0.1801 mol

[N2] = 0.1801 mol / 4.50 L = 0.04002 mol/L

H2: 1.661 mol - 0.843 mol = 0.818 mol

[H2] = 0.818 mol / 4.50 L = 0.182 mol/L

NH3: 0.562 mol

[NH3] = 0.562 mol / 4.50 L = 0.125 mol/L

Step 6: Calculate Kc

Kc = [NH3]² / ([N2] [H2]³)

Kc = 0.125² /(0.04002*0.182³)

Kc = 64.76

The equilibrium constant is 64.76