JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
At \(298\,K\), the equilibrium constant is \(2 \times 10^{15}\) for the reaction. \(Cu ( s )+2 Ag ^{+}( aq ) \rightleftharpoons Cu ^{2+}( aq )+2 Ag ( s )\) The equilibrium constant for the reaction.\(\frac{1}{2} Cu ^{2+}( aq )+ Ag ( s ) \rightleftharpoons \frac{1}{2} Cu ( s )+ Ag ^{+}( aq )\) is \(x \times 10^{-8}\). The value of \(x\) is (Nearest Integer)
- A \(3\)
- B \(0\)
- C \(1\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(K_{e q}^{\prime}=\frac{1}{\sqrt{K_{e q}}}=\frac{1}{\sqrt{2 \times 10^{15}}}=x \times 10^{-8}\) \(\frac{1}{\sqrt{20}} \times \frac{1}{10^{7}}= x \times 10^{-8}\) \(\frac{1}{\sqrt{20}} \times 10^{-7}= x \times 10^{-8}\) \(\frac{10}{\sqrt{20}}= x\)…
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