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 $\frac{-\mathrm{d}\left[\left[\mathrm{PtCl}_{4}\right]^{2-}\right]}{\mathrm{dt}}=4.8 \times 10^{-5}\left[\left[\mathrm{PtCl}_{4}\right]^{2-}\right]-2.4 \times 10^{-3}\left[\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]^{-}\right]\left[\mathrm{Cl}^{-}\right]$ where square brackets are used to denote molar concentrations. The equilibrium constant $\mathrm{K}_{\mathrm{C}}=$ $\_\_\_\_$. (Nearest integer)

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Chemical Kinetics

The reaction rate for the reaction

[PtCl_{4}]^{2 -} + H_{2} O ⇌ [Pt (H_{2} O) Cl_{3}]^{-} + Cl^{-}

was measured as a function of concentrations of different species. It was observed that

\frac{- d [ [ PtCl _{4} ] ^{2 -} ]}{dt} = 4.8 \times 1 0^{- 5} [[PtCl_{4}]^{2 -}] - 2.4 \times 1 0^{- 3} [[Pt (H_{2} O) Cl_{3}]^{-}] [Cl^{-}]

where square brackets are used to denote molar concentrations. The equilibrium constant

K_{C} =

____

. (Nearest integer)

Answer: 0

At equilibrium the net rate is zero, so the forward term equals the reverse term:

4.8 \times 1 0^{- 5} [PtCl_{4}^{2 -}] = 2.4 \times 1 0^{- 3} [Pt(H_{2} O)Cl_{3}^{-}] [Cl^{-}]

. Here

4.8 \times 1 0^{- 5}

is the forward rate constant

k_{f}

and

2.4 \times 1 0^{- 3}

the reverse

k_{b}

, so

K_{C} = \frac{[ Pt(H _{2} O)Cl _{3}^{-} ] [ Cl ^{-} ]}{[ PtCl _{4}^{2 -} ]} = \frac{k _{f}}{k _{b}} = \frac{4.8 \times 1 0 ^{- 5}}{2.4 \times 1 0 ^{- 3}} = 0.02

. To the nearest integer,

K_{C} = 0

. (The value 50 corresponds to

k_{b} / k_{f}

, the constant for the reverse reaction, not the forward reaction as written.)

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer