TS EAMCET · Chemistry · Chemical Equilibrium
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 \(\mathrm{T}(\mathrm{K})\) ? \(\mathrm{A}(\mathrm{~g})+\frac{1}{2} \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{D}(\mathrm{~g})\)
- A \(100\)
- B \(50\)
- C \(20\)
- D \(75\)
Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
\(2 \mathrm{~A}(\mathrm{~g}) \therefore \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}) ; \mathrm{K}_1=16\) \(\mathrm{K}_1=\frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2}=16\) ...(i)…
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