Question 20 of 69intermediate🔧 ApplyNumerical3 marks

One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2_2.

FeO(s)+CO(g)⇌Fe(s)+CO2(g);Kp=0.265 atm at 1050 KFeO (s) + CO (g) \rightleftharpoons Fe (s) + CO_2(g); \quad K_p = 0.265 \text{ atm at 1050 K}

What are the equilibrium partial pressures of CO and CO2_2 at 1050 K if the initial partial pressures are: pCO=1.4p_{CO} = 1.4 atm and pCO2=0.80p_{CO_2} = 0.80 atm?

Correct Answer

Equilibrium partial pressure of CO is 1.739 atm and CO2_2 is 0.461 atm.

Exercise: EXERCISES | Q: 6.20 | (Page 43)
For More Understanding

Explanation

For the given heterogeneous equilibrium, pure solids FeO and Fe do not appear in the equilibrium constant expression. Thus, KpK_p depends only on the partial pressures of the gaseous components, CO and CO2_2. By setting up an ICE table equivalent with an unknown change xx, we can solve for the equilibrium partial pressures. The negative value of xx indicates that the reaction actually shifts in the backward direction to reach equilibrium, increasing the partial pressure of CO and decreasing that of CO2_2.

Solution Steps

  1. Step 1: Write the expression for the equilibrium constant. Since FeO(s) and Fe(s) are pure solids, their partial pressures are not included. Thus, Kp=pCO2pCOK_p = \frac{p_{CO_2}}{p_{CO}}.

  2. Step 2: Define the equilibrium partial pressures in terms of xx. Let the reaction proceed in the forward direction by xx atm. At equilibrium: pCO=1.4−xp_{CO} = 1.4 - x and pCO2=0.80+xp_{CO_2} = 0.80 + x.

  3. Step 3: Substitute the equilibrium partial pressures into the KpK_p expression. 0.265=0.80+x1.4−x0.265 = \frac{0.80 + x}{1.4 - x}.

  4. Step 4: Solve for xx. 0.80+x=0.265(1.4−x)⇒0.80+x=0.371−0.265x⇒1.265x=−0.429⇒x=−0.3390.80 + x = 0.265(1.4 - x) \Rightarrow 0.80 + x = 0.371 - 0.265x \Rightarrow 1.265x = -0.429 \Rightarrow x = -0.339 atm.

  5. Step 5: Calculate the final equilibrium partial pressures. The negative value of xx means the reaction shifts backward. pCO=1.4−(−0.339)=1.739p_{CO} = 1.4 - (-0.339) = 1.739 atm and pCO2=0.80+(−0.339)=0.461p_{CO_2} = 0.80 + (-0.339) = 0.461 atm.

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