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JEE Main 202126 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2021 · 26 August · Evening Shift · Chemistry · Chemical Equilibrium

The reaction rate for the reaction [ PtCl _ 4 ] ^ 2 - + H _ 2 O ⇌...

The reaction rate for the reaction

[PtCl4]2−+H2O⇌[Pt(H2O)Cl3]−+Cl−{\left[{PtCl}_{4}\right]}^{2 -} + {H}_{2}O ⇌ {\left[Pt\left({H}_{2}O\right){Cl}_{3}\right]}^{-} + {Cl}^{-}

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

−d[[PtCl4]2−]dt=4.8×10−5[[PtCl4]2−]−2.4×10−3[[Pt(H2O)Cl3]−][Cl−]\dfrac{- d\left[{\left[{PtCl}_{4}\right]}^{2 -}\right]}{dt} = 4 . 8 \times {10}^{- 5}\left[{\left[{PtCl}_{4}\right]}^{2 -}\right] - 2 . 4 \times {10}^{- 3}\left[{\left[Pt\left({H}_{2}O\right){Cl}_{3}\right]}^{-}\right]\left[{Cl}^{-}\right]

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

Kc=X{K}_{c} = X(Nearest integer)

Value of 1Xis\dfrac{1}{X}is (question is modified.)

Correct answer

0

Step-by-step solution

KC=KfKb=4.8×10−52.4×10−3=2×10−2Valueof1Kc=50{K}_{C} = \dfrac{{K}_{f}}{{K}_{b}} = \dfrac{4 . 8 \times {10}^{- 5}}{2 . 4 \times {10}^{- 3}} = 2 \times {10}^{- 2}Valueof\dfrac{1}{{K}_{c}} = 50

Note: This question is modified from the original.

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