🛠️ JEE🧪 Chemistry

The reaction rate for the reaction \[\left[\mathrm{PtCl}_{4}\right]^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\le…

Q1

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 concentration of different species. It was observed that

\[\begin{array}{l}\frac{-\mathrm{d}\left[\left[\mathrm{PtCl}_{4}\right]^{2-}\right]}{\mathrm{dt}}=4.8 \times 10^{-5} \\ {\left[\left[\mathrm{PtCl}_{4}\right]^{2-}\right]-2.4 \times 10^{-3}\left[\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]^{-}\right]\left[\mathrm{Cl}^{-}\right] \text {. }} \\\end{array}\]

where square brackets are used to denote molar concentration. The equilibrium constant \( \mathrm{K}_{\mathrm{C}}= \) _____\(\text{x}{10}^{-2}\). (Nearest integer)

a

1

b

2

c

3

d

4

🔒
Answer & explanation — PYQ Pass

Options are free to see. Unlock the correct answer and full explanation with Pass.

Unlock · ₹149

Practice more JEE Chemistry PYQs

See every question on Equilibrium, or browse the full JEE question bank.

See all questions on Equilibrium →