JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The reaction rate for the reaction \(\left[\mathrm{PtCl}_{4}\right]^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]^{-}+\mathrm{Cl}^{-}\) was measured as a function of concentrations of different species. It was observed that \(\frac{-\mathrm{d}\left[\left[\mathrm{PtCl}_{4}\right]^{2-}\right]}{\mathrm{dt}}=4.8 \times 10^{-5}\left[\left[\mathrm{PtCl}_{4}\right]^{2-}\right]-2.4 \times10^{-3}\)
\(\left[\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]^{-}\right]\left[\mathrm{Cl}^{-}\right]\) where square brackets are used to denote molar concentrations. The equilibrium constant \(\mathrm{K}_{\mathrm{c}}=....\). (Nearest integer)
- A \(500\)
- B \(50\)
- C \(5\)
- D \(0.5\)
Answer & Solution
Correct Answer
(B) \(50\)
Step-by-step Solution
Detailed explanation
\({\left[\mathrm{PtCl}_{4}\right]^{-2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]^{-}+\mathrm{Cl}^{-}}\)…
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