Consider the equilibrium system described by the chemical reaction below. At equilibrium, a 2.0 L reaction vessel contained a mixture of 1.2 mol Fe, 1.0 × 10-³ mol O2, and 2.0 mol of Fe2O, at 2000.0 °C. What are the values of Kc and Kp for this reaction? 4Fe(s) + 302(g) = 2Fe2O2(s) NEXT Based on the given data, set up the expression for Kc. Each reaction participant must be represented by one tile. Do not combine terms. [1.2] Once the expression is constructed, solve for Kc. Complete Parts 1-2 before submitting your answer. 2[2.0] 2[1.2] [1.0 × 10³] [0.50] 8.0 × 10° ?? 1 = [2.0] [5.0 x 10²] 1.0 × 10? [1.2]¹ [0.50]* 1.3 x 10-1° [1.0 × 10.³1³ [5.0 × 10.¹1³ 2 = [2.0]² [1.2]² 4[1.2] 4[0.50] > RESET 3[1.0 × 10³] 3[5.0 x 10+]
The correct answer and explanation is:
Step 1: Write the Expression for KcK_c
The equilibrium constant KcK_c is expressed as: Kc=[Products]coefficients[Reactants]coefficientsK_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}
For the given reaction: 4Fe(s)+3O2(g)⇌2Fe2O2(s)4Fe(s) + 3O_2(g) \rightleftharpoons 2Fe_2O_2(s)
Since solids (FeFe and Fe2O2Fe_2O_2) are not included in the equilibrium expression: Kc=[Fe2O2]2[O2]3K_c = \frac{[Fe_2O_2]^2}{[O_2]^3}
Using the given equilibrium concentrations:
- [Fe2O2]=2.0 mol2.0 L=1.0 M[Fe_2O_2] = \frac{2.0 \text{ mol}}{2.0 \text{ L}} = 1.0 \text{ M}
- [O2]=1.0×10−3 mol2.0 L=5.0×10−4 M[O_2] = \frac{1.0 \times 10^{-3} \text{ mol}}{2.0 \text{ L}} = 5.0 \times 10^{-4} \text{ M}
Substituting these values: Kc=(1.0)2(5.0×10−4)3K_c = \frac{(1.0)^2}{(5.0 \times 10^{-4})^3} Kc=1.0(1.25×10−10)K_c = \frac{1.0}{(1.25 \times 10^{-10})} Kc=8.0×109K_c = 8.0 \times 10^9
Step 2: Calculate KpK_p
The relationship between KpK_p and KcK_c is: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}
where:
- R=0.0821R = 0.0821 L·atm/(mol·K) (Ideal gas constant)
- T=2000.0T = 2000.0 K (given temperature)
- Δn=\Delta n = (moles of gas products – moles of gas reactants)
- Products: 00 (since Fe₂O₂ is solid)
- Reactants: 33 (from O₂ gas)
- Δn=0−3=−3\Delta n = 0 – 3 = -3
Kp=(8.0×109)×(0.0821×2000)−3K_p = (8.0 \times 10^9) \times (0.0821 \times 2000)^{-3} Kp=(8.0×109)×(164.2)−3K_p = (8.0 \times 10^9) \times (164.2)^{-3} Kp=(8.0×109)×(2.24×10−6)K_p = (8.0 \times 10^9) \times (2.24 \times 10^{-6}) Kp=1.79×104K_p = 1.79 \times 10^4
Final Answer:
- Kc=8.0×109K_c = 8.0 \times 10^9
- Kp=1.79×104K_p = 1.79 \times 10^4
This indicates that the equilibrium heavily favors the formation of Fe2O2Fe_2O_2 over oxygen gas.
Now, generating an image illustrating the chemical equilibrium:
Here is a scientific illustration of the chemical equilibrium reaction at 2000°C, showing the balance between solid iron (Fe), gaseous oxygen (O₂), and iron oxide (Fe₂O₂) in a reaction vessel. Let me know if you need any modifications!
