WBJEE · Chemistry · Chemical Equilibrium
The equilibrium constant for the following reactions are given at \(25^{\circ} \mathrm{C}\)
\(\begin{array}{l}
2 \mathrm{~A} \rightleftharpoons \mathrm{B}+\mathrm{C}, \mathrm{K}_{1}=1.0 \\
2 \mathrm{~B} \rightleftharpoons \mathrm{C}+\mathrm{D}, \mathrm{K}_{2}=16 \\
2 \mathrm{C}+\mathrm{D} \rightleftharpoons 2 \mathrm{P}, \mathrm{K}_{3}=25
\end{array}\)
The equilibrium constant for the reaction \(\mathrm{P} \rightleftharpoons \mathrm{A}+\frac{1}{2} \mathrm{~B}\) at \(25^{\circ} \mathrm{C}\) is
- A \(\frac{1}{20}\)
- B 20
- C \(\frac{1}{42}\)
- D 21
Answer & Solution
Correct Answer
(A) \(\frac{1}{20}\)
Step-by-step Solution
Detailed explanation
Hint: We can manipulate the given equations as follows \(\frac{1}{2} \mathrm{~B}+\frac{1}{2} \mathrm{C} \rightleftharpoons \mathrm{A}, \quad \mathrm{K}_{1}^{\prime}=1\)…
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