NEET 2024 — Chemistry solution

NEET2024Chemistry

Consider the following reaction in a sealed vessel at equilibrium with concentrations of N $_{2} = 3.0 \times 1 0^{- 3}$ M, O $_{2} = 4.2 \times 1 0^{- 3}$ M and NO $= 2.8 \times 1 0^{- 3}$ M. $2 NO (g) ⇌ N_{2} (g) + O_{2} (g)$ . If 0.1 mol/L of NO(g) is taken in a closed vessel, what will be the degree of dissociation ( $α$ ) of NO(g) at equilibrium?

A
0.8889
B
0.717
C
0.00889
D
0.0889

Solution

From the given equilibrium concentrations: $K_{c} = \frac{[ N _{2} ] [ O _{2} ]}{[ NO ] ^{2}} = \frac{( 3.0 \times 1 0 ^{- 3} ) ( 4.2 \times 1 0 ^{- 3} )}{( 2.8 \times 1 0 ^{- 3} ) ^{2}} \approx 1.607$ .

Starting with $[NO] = 0.1$ M and letting fraction $α$ dissociate, at equilibrium $[NO] = 0.1 (1 - α)$ , $[N_{2}] = [O_{2}] = 0.05 α$ .

$K_{c} = \frac{( 0.05 α ) ^{2}}{[ 0.1 ( 1 - α ) ] ^{2}} = \frac{α ^{2}}{4 ( 1 - α ) ^{2}} = 1.607$ .

Taking square roots: $\frac{α}{2 ( 1 - α )} = 1.268 \Rightarrow α = 2.536 - 2.536 α \Rightarrow α \approx 0.717$ .

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