In a chemical reaction A + 2B K 2C + D, the initial concentration of B was 1.5 times of the concentration of A, but the…

Step 1 — Set up variables:

Let initial $[\mathrm{A}] = a$, initial $[\mathrm{B}] = 1.5a$, initial $[\mathrm{C}] = [\mathrm{D}] = 0$.

Let $x$ = moles of A reacted per litre.

Step 2 — ICE table:

| | A | B | C | D | |--|--|--|--|--| | Initial | $a$ | $1.5a$ | 0 | 0 | | Change | $-x$ | $-2x$ | $+2x$ | $+x$ | | Equil. | $a-x$ | $1.5a-2x$ | $2x$ | $x$ |

Step 3 — Use condition $[\mathrm{A}]_{eq} = [\mathrm{B}]_{eq}$:

$a - x = 1.5a - 2x \Rightarrow x = 0.5a$

Step 4 — Equilibrium concentrations:

$[\mathrm{A}] = [\mathrm{B}] = a - 0.5a = 0.5a$

$[\mathrm{C}] = 2(0.5a) = a, \quad [\mathrm{D}] = 0.5a$

Step 5 — Calculate K:

$K = \frac{[\mathrm{C}]^2[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]^2} = \frac{a^2 \cdot 0.5a}{0.5a \cdot (0.5a)^2} = \frac{0.5a^3}{0.5a \cdot 0.25a^2} = \frac{0.5a^3}{0.125a^3} = 4$

Answer: Option (2) — K = 4