The reaction, \(2 \mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \leftrightharpoons 3…

The reaction,
\(2 \mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \leftrightharpoons 3 \mathrm{C}_{(\mathrm{g})}+\mathrm{D}_{(\mathrm{g})}\)
is begun with the concentrations of \(A\) and \(B\) both at an initial value of 1.00 M When equilibrium is reached, the concentration of \(D\) is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
  1. \(\left[(0.75)^3(0.25)\right] \div\left[(1.00)^2(1.00)\right]\)
  2. \(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.75)\right]\)
  3. \(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.25)\right]\)
  4. \(\left[(0.75)^3(0.25)\right] \div\left[(0.75)^2(0.25)\right]\)

Solution

$\begin{aligned} 2 A_{(g)} & + B_{(g)} & \rightleftharpoons & 3 C_{(g)} & + D_{(g)} \\ \text{Initial moles : } & 1 & & 1 & & 0 & & 0 \\ \text{Moles at eq. : } & 1-(2 \times 0.25) & & 1-0.25 & & 3 \times 0.25 & & 0.25 \\ & =0.5 & & =0.75 & & =0.75 & & =0.25 \end{aligned}$ Equilibrium constant, \(K=\frac{[C]^3[D]}{[A]^2[B]}\)
\(\therefore \quad K=\frac{(0.75)^3(0.25)}{(0.5)^2(0.75)}\)

Asked in: NEET 2010 (Mains)

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