The equilibrium constant for the decomposition of H₂O(g) ⇌ H₂(g) + ½O₂(g) (ΔG° = 92.34 kJ/mol) is 8.0 × 10⁻³ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation (α) of water is ______ ×10⁻²(nearest integer value).

 Question:

The equilibrium constant for the decomposition of

$$H_{2}O(g)\rightleftharpoons H_2(g)+\frac{1}{2}{O}_2(g)$$

is K = 8.0 × 10⁻³ at 2300 K.
If the total equilibrium pressure = 1 bar, the degree of dissociation (α) of water is ____ × 10⁻² (nearest integer).

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Short Solution (Text):

Step 1: Initial moles

Take 1 mole of H₂O initially.

At equilibrium (α = degree of dissociation):

• H₂O = (1 − α)

• H₂ = α

• O₂ = α/2

Total moles = $1 − α + α + α/2 = 1 + α/2$

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Step 2: Mole fractions

$$y_{H_{2}O}=\frac{1-\alpha }{1+\alpha \mathrm{/}2},y_{H_2}=\frac{\alpha }{1+\alpha \mathrm{/}2},y_{O_2}=\frac{\alpha \mathrm{/}2}{1+\alpha \mathrm{/}\mathrm{2}}$$

Since total pressure = 1 bar, partial pressure = mole fraction × 1 bar.

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Step 3: Expression for $K_p$

$$K_p=\frac{(p_{H_2})(p_{O_2}{)}^{1\mathrm{/}2}}{p_{H_2\mathrm{O}}}$$

Substitute:

$$K_p=\frac{\frac{\alpha }{1+\alpha \mathrm{/}2}\cdot \sqrt{\frac{\alpha \mathrm{/}2}{1+\alpha \mathrm{/}2}}}{\frac{1-\alpha }{1+\alpha \mathrm{/}\mathrm{2<span\; class="vlist-s"></span>}}}$$

Simplify (denominators cancel):

$$K_p = \frac{α \cdot \sqrt{α/2}}{(1 − α)\sqrt{1 + α/2}}$$

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Step 4: Approximation (since K is small, α ≪ 1)

For small α: $1 − α ≈ 1$, $\sqrt{1+α/2} ≈ 1$

So,

$$K_p ≈ α \cdot \sqrt{\frac{α}{2}} = \frac{α^{3/2}}{\sqrt{2}}$$

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Step 5: Solve for α

$$α^{3/2} = K_p \cdot \sqrt{2} = (8.0 \times 10^{-3})(1.414) \approx 1.13 \times 10^{-2}$$
$$\alpha =(1.13\times {10}^{-2}{)}^{2\mathrm{/}3}$$

Take log or approximate:

• $10^{-2}$ to power $2/3$ ≈ $10^{-1.33} \approx 0.0467$
  

• Multiply factor (1.13)^{2/3} ≈ 1.085

$$\alpha \approx 0.0507\approx 5.1\times {10}^{-2}$$

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✅ Final Answer:
Degree of dissociation = 5 × 10⁻² (nearest integer).

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Conclusion – Video Solution:

By applying the equilibrium expression in terms of α, simplifying under the low dissociation approximation, and solving for α, we find the degree of dissociation of water = ≈ 5 × 10⁻².