Question 31 of 69intermediate🔧 ApplyNumerical3 marks

Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2_2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pco=pH2O=4.0p_{co} = p_{H_2O} = 4.0 bar, what will be the partial pressure of H2_2 at equilibrium?

Kp=10.1 at 400∘CK_p = 10.1 \text{ at } 400^\circ C
Correct Answer

The partial pressure of H2H_2 at equilibrium is 3.043.04 bar.

Exercise: EXERCISES | Q: 6.31 | (Page 45)
For More Understanding

Explanation

To find the equilibrium partial pressure of H2, we set up an ICE table in terms of partial pressures. Since the reaction starts with equimolar amounts of CO and H2O, and produces equal moles of CO2 and H2, the equilibrium expression simplifies to a ratio of squares. By substituting the given Kp value and initial pressures, we can solve the resulting quadratic equation for the change in pressure, which directly gives the partial pressure of H2.

Solution Steps

  1. Step 1: Write the reaction and initial partial pressures.

    CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)

    Initial pressures: pCO=4.0p_{CO} = 4.0 bar, pH2O=4.0p_{H_2O} = 4.0 bar, pCO2=0p_{CO_2} = 0, pH2=0p_{H_2} = 0.

  2. Step 2: Let the partial pressure of H2H_2 at equilibrium be xx bar. Then, pCO2=xp_{CO_2} = x bar, and the remaining reactant pressures will be pCO=pH2O=(4.0−x)p_{CO} = p_{H_2O} = (4.0 - x) bar.

  3. Step 3: Write the expression for KpK_p and substitute the equilibrium pressures.

    Kp=pCO2×pH2pCO×pH2O=x×x(4.0−x)(4.0−x)=x2(4.0−x)2K_p = \frac{p_{CO_2} \times p_{H_2}}{p_{CO} \times p_{H_2O}} = \frac{x \times x}{(4.0 - x)(4.0 - x)} = \frac{x^2}{(4.0 - x)^2}

  4. Step 4: Substitute Kp=10.1K_p = 10.1 and solve for xx.

    10.1=x2(4.0−x)210.1 = \frac{x^2}{(4.0 - x)^2}

    Taking the square root of both sides:

    10.1=x4.0−x\sqrt{10.1} = \frac{x}{4.0 - x} 3.178=x4.0−x3.178 = \frac{x}{4.0 - x} 3.178(4.0−x)=x3.178(4.0 - x) = x 12.712−3.178x=x12.712 - 3.178x = x 12.712=4.178x12.712 = 4.178x

    x=12.7124.178≈3.04x = \frac{12.712}{4.178} \approx 3.04 bar.

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