At temperature T, a compound `AB_2(g)` dissociation according to the reaction, `2AB _2(g) hArr 2AB(g) +B_2(g)` with degree of dissociation, `alpha`, which is small compared to unity . Deduce the expression for `alpha` in terms of the equilibrium constant `K_p` and the total pressure P.

Explanation

To deduce the expression for the degree of dissociation (α) of the compound \( AB_2(g) \) in terms of the equilibrium constant \( K_p \) and the total pressure \( P \), we will follow these steps:

### Step 1: Write the Reaction and Initial Conditions
The dissociation reaction is given as:
\[ 2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g) \]
Initially, we have 2 moles of \( AB_2 \) and no products:
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