Q10 / 160 · TS EAMCET 2024 (11 May Shift 1) · Chemistry
At $T(K)$ the equilibrium constants for the following two reactions are given below
$\begin{aligned}
& 2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{~g}) ; \mathrm{K}_1=16 \\
& 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{D}(\mathrm{~g}) ; \mathrm{K}=25
\end{aligned}$
What is the value of equilibrium constant ( K ) for the reaction given below at $\mathrm{T}(\mathrm{K})$ ?
$\mathrm{A}(\mathrm{~g})+\frac{1}{2} \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{D}(\mathrm{~g})$
What is the value of equilibrium constant ( K ) for the reaction given below at $\mathrm{T}(\mathrm{K})$ ? $\mathrm{A}(\mathrm{~g})+\frac{1}{2} \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{D}(\mathrm{~g})$
- A$100$
- B$50$
- Correct answerC$20$
- D$75$
Solution
$2 \mathrm{~A}(\mathrm{~g}) \therefore \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}) ; \mathrm{K}_1=16$
$\mathrm{K}_1=\frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2}=16$ ...(i)
$2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \therefore 2 \mathrm{D}(\mathrm{g}) ; \mathrm{K}_2=25$
$\mathbb{K}_2=\frac{[\mathrm{D}]^2}{[\mathrm{C}][\mathrm{B}]^2}=25$ ...(ii)
$\therefore \quad \mathrm{A}+\frac{1}{2} \mathrm{~B}(\mathrm{~g}) \therefore \mathrm{D} ; \mathrm{K}_3$
$\mathrm{K}_3=\frac{[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]^{1 / 2}}$ ...(iii)
$\begin{aligned} & \mathbb{K}_3=\sqrt{\mathbb{K}_2 \times \mathbb{K}_1} \\ & =\sqrt{\frac{[\mathrm{D}]^2}{[\mathrm{C}][\mathrm{B}]^2} \times \frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2}}=\sqrt{25 \times 16} \\ & \Rightarrow \frac{[\mathrm{D}]}{[\mathrm{B}]^{1 / 2}[\mathrm{~A}]}=4 \times 5=20\end{aligned}$
Hence, the correct answer is option: 'C'.