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Degree of Dissociation Using Equilibrium Constant

Learn how to calculate degree of dissociation using equilibrium constant and total pressure for gaseous reactions. This method helps solve JEE...

❓ Question

For decomposition of steam:

H2O(g)H2(g)+12O2(g)H_2O(g)\rightleftharpoons H_2(g)+\frac12 O_2(g)
Kp=8.0×103K_p=8.0\times10^{-3}

at:

2300 K2300\text{ K}

Total pressure at equilibrium:

1 bar1\text{ bar}

Degree of dissociation of water is:

α=____×102\alpha=\_\_\_\_\times10^{-2}

(Nearest integer value)

Assume:

α1\alpha \ll 1

đź–Ľ Question Image

Degree of Dissociation Using Equilibrium Constant


✍️ Short Explanation

This problem is based on:

👉 Degree of dissociation
👉 Equilibrium constant KpK_p
👉 Partial pressures.

Main idea:

For decomposition:

H2O(g)H2(g)+12O2(g)H_2O(g)\rightleftharpoons H_2(g)+\frac12 O_2(g)

if initial moles of water are 1:

Kp=PH2PO21/2PH2O\boxed{ K_p= \frac{P_{H_2}\cdot P_{O_2}^{1/2}} {P_{H_2O}} }

Degree of Dissociation Using Equilibrium Constant

đź”· Step 1 — Assume Initial Moles đź’Ż

Take initially:

1 mole H2O1\text{ mole } H_2O

Let degree of dissociation be:

α\alpha

Reaction:

H2OH2+12O2H_2O \rightleftharpoons H_2+\frac12 O_2

Equilibrium Moles

SpeciesInitialChangeEquilibrium
H2OH_2O
1α-\alpha
1α1-\alpha
H2H_20+α+\alpha
α\alpha
O2O_20+α2+\frac{\alpha}{2}α2\frac{\alpha}{2}

đź”· Step 2 — Total Moles at Equilibrium

ntotal=1α+α+α2n_{total} = 1-\alpha+\alpha+\frac{\alpha}{2}
=1+α2= 1+\frac{\alpha}{2}

Given:

α1\alpha\ll1

So:

ntotal1n_{total}\approx1

đź”· Step 3 — Partial Pressures

Total pressure:

P=1 barP=1\text{ bar}

Thus:

PH2αP_{H_2}\approx\alpha
PO2α2P_{O_2}\approx\frac{\alpha}{2}
PH2O1P_{H_2O}\approx1

đź”· Step 4 — Apply KpK_p Expression

Kp=PH2PO21/2PH2OK_p= \frac{P_{H_2}\cdot P_{O_2}^{1/2}} {P_{H_2O}}

Substitute:

8×103=α(α2)1/28\times10^{-3} = \alpha\left(\frac{\alpha}{2}\right)^{1/2}
=α3/22= \frac{\alpha^{3/2}}{\sqrt2}

Thus:

α3/2=8×103×2\alpha^{3/2} = 8\times10^{-3}\times\sqrt2
1.13×102\approx1.13\times10^{-2}

đź”· Step 5 — Solve for α\alpha

α=(1.13×102)2/3\alpha = (1.13\times10^{-2})^{2/3}
0.050\approx0.050

Thus:

α5.0×102\alpha\approx5.0\times10^{-2}

Nearest integer:

5\boxed{5}

đź”· Step 6 — JEE Trap Alert 🚨

❌ Total moles calculate karte time:

α2\frac{\alpha}{2}

bhool jaana

❌ Partial pressure directly mole maan lena without total pressure

KpK_p expression galat likh dena

Remember:

Pi=nintotalPtotal\boxed{ P_i= \frac{n_i}{n_{total}}P_{total} }

✅ Final Answer

5\boxed{5}

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