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

Question

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, $\left[(0.75)^3(0.25)\right] \div\left[(1.00)^2(1.00)\right]$, $\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.75)\right]$, $\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.25)\right]$, $\left[(0.75)^3(0.25)\right] \div\left[(0.75)^2(0.25)\right]$

MARKS App · Accepted Answer

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

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