AP EAMCET · Chemistry · Chemical Equilibrium
At \(\mathrm{T}(\mathrm{K})\), the equilibrium constant for the reaction \(\mathrm{a} \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{b} \mathrm{B}(\mathrm{g})\) is \(\mathrm{K}_{\mathrm{c}}\). If the reaction takes place in the following form \(2 \mathrm{aA}(\mathrm{g}) \rightleftharpoons 2 \mathrm{bB}(\mathrm{g})\) its equilibrium constant is \(\mathrm{K}_{\mathrm{c}}^{\prime}\). The correct relationship between \(\mathrm{K}_{\mathrm{c}}\) and \(\mathrm{K}_{\mathrm{c}}\) is
- A \(\mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^2\)
- B \(\mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^{\frac{1}{2}}\)
- C \(\mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^{-1}\)
- D \(\mathrm{K}_{\mathrm{c}}^{\prime}=\mathrm{K}_{\mathrm{c}}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{K}_{\mathrm{c}}^{\prime}=\left(\mathrm{K}_{\mathrm{c}}\right)^2\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {} \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{B}]^{\mathrm{b}}}{[\mathrm{A}]^{\mathrm{a}}} \text { and } \mathrm{K}_{\mathrm{c}}^{\prime}=\frac{[\mathrm{B}]^{2 \mathrm{~b}}}{[\mathrm{~A}]^{2 \mathrm{a}}} \\ & \Rightarrow…
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