Q38 / 120 · NEET 2010 (Mains) · Chemistry

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

  • A
    \(\left[(0.75)^3(0.25)\right] \div\left[(1.00)^2(1.00)\right]\)
  • Correct answer
    B
    \(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.75)\right]\)
  • C
    \(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.25)\right]\)
  • D
    \(\left[(0.75)^3(0.25)\right] \div\left[(0.75)^2(0.25)\right]\)

Solution

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

Hence, the correct answer is option: 'B'.

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