Question

Consider the following reaction at 25 °C: 4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s) An equilibrium mixture contains 1.0 mol Fe, 1.0 × 10–3 mol O2, and 2.0 mol of Fe2O3 all in a 2.0 L container. Calculate the value of K for this reaction. What is the value of Kp for this reaction. (Answer: 8.0 × 109 M–3; 5.5 × 105 atm–3)

Answer 1

Answer:

Kc = Kc = 8.0 * 10^9

Kp = 5.5 *10^5

Explanation:

Step 1: Data given

Temperature = 25.0 °C

Number of moles Fe = 1.0 moles

Number of moles O2 = 1.0 * 10^-3 moles

Number of moles Fe2O3 = 2.0 moles

Volume = 2.0 L

Step 2: The balanced equation

4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s)

Step 3: Calculate molarity

Molarity = moles / volume

[Fe] = 1.0 moles / 2.0 L

[Fe] = 0.5 M

[O2] = 0.001 moles / 2.0 L

[O2] = 0.0005 M

[Fe2O3] = 2.0 moles / 2.0 L

[Fe2O3] = 1.0 M

Step 4: Calculate Kc

Kc =1/ [O2]³

Kc = 1/0,.000000000125

Kc = 8.0 * 10^9

Step 5: Calculate Kp

Kp = Kc*(R*T)^Δn

⇒with Kc = 8.0*10^9

⇒with R = 0.08206 L*atm /mol*K

⇒with T = 298 K

⇒with Δn = -3

Kp = 8.10^9 *(0.08206 * 298)^-3

Kp = 5.5 *10^5