The approach of the following equilibrium was observed kinetically from both directions PtCl 42- H 2 O - Pt H 2 O Cl 3- Cl -At 25- C- it is found - Idfrac - d - PtCl - 4 -2- dt -3-9 - 10-5 s -1 - PtCl - 4 -2-2-1 - 10-3Lmol-1 s -1- Pt H - 2 O Cl - 3- Cl - sValue of K C when fourth Cl -is complexed in the reaction is-A- 1-85 - 10-2B- 1-86 - 2C- 53-85D- 8-19 - 10-8

The correct option is C 53.85For the given reaction, k_f=3.9×10^ 5 s^ 1 k_b=2.1× 10^ 3 Lmol^ 1s^ 1 ∴ K_c=k_f/k_b=3.9×10^ 5/2.1× 10^ 3 molL^ 1=0.0186 ∴ K_c^/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Effective Equilibrium Constant

The approach ...

Question

The approach of the following equilibrium was observed kinetically from both directions
$P t C l_{4}^{2 -} + H_{2} O ⇌ P t (H_{2} O) C l_{3}^{-} + C l^{-}$
At 25°C, it is found $\frac{- d [P t C l_{4}^{2 -}]}{d t} = (3.9 \times 10^{- 5} s^{- 1)}) [P t C l_{4}^{2 -}] - (2.1 \times 10^{- 3} L m o l^{- 1} s^{- 1}) [P t (H_{2} O) C l_{3}^{-}] [C l^{-}]$
Value of $K_{C}$ when fourth $C l^{-}$ is complexed in the reaction is:

1.85 \times 10^{- 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.86 \times 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

53.85

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8.19 \times 10^{- 8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $53.85$
For the given reaction,
$k_{f} = 3.9 \times 10^{- 5} s^{- 1}$
$k_{b} = 2.1 \times 10^{- 3} L m o l^{- 1} s^{- 1}$

$∴ K_{c} = \frac{k_{f}}{k_{b}} = \frac{3.9 \times 10^{- 5}}{2.1 \times 10^{- 3}} m o l L^{- 1} = 0.0186$

$∴ K_{c}^{/}$ of the required reaction $= \frac{1}{K_{c}} = 53.85$

Suggest Corrections

Similar questions

Q. The approach of the following equilibrium was observed kinetically from both directions

P t C l_{4}^{2 -} + H_{2} O ⇌ P t (H_{2} O) C l_{3}^{-} + C l^{-}

At 25°C, it is found

\frac{- d [P t C l_{4}^{2 -}]}{d t} = (3.9 \times 10^{- 5} s^{- 1)}) [P t C l_{4}^{2 -}] - (2.1 \times 10^{- 3} L m o l^{- 1} s^{- 1}) [P t (H_{2} O) C l_{3}^{-}] [C l^{-}]

Value of

K_{C}

when fourth

C l^{-}

is complexed in the reaction is:

Q. What is

p K

value of the complex

[P t C l_{4}]

if__________.

P t^{4 +} + C l^{-} \to [P t C l]^{3 +}

K_{1} = 10^{- 2}

[P t C l]^{3 +} + C l^{-} \to [P t C l_{2}]^{2 +}

K_{2} = 10^{- 3}

[P t C l_{2}]^{2 +} + C l^{-} \to [P t C l_{3}]^{+}

K_{3} = 10^{- 4}

[P t C l_{3}]^{+} + C l^{-} \to [P t C l_{4}]

K_{4} = 10^{- 4}

Give the iupac names

[Pt(NH3)4][PtCl4]

[Pt(NH3)4Cl2][PtCl4]

A sample of pure $P C l_{5}$ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl_5 was found to be $0.5 \times 10^{- 1} m o l L^{- 1} . I f v a l u e o f K_{c} i s 8.3 \times 10^{- 3}$ , what are the concentrations of $P C l_{3}$ and $C l_{2}$ at equilibrium?\\
$P C l_{5} (g) \leftrightarrow P C l_{3} (g) + C l_{2} (g)$