The approach of the following equilibrium was observed kinetically from both directionsPtCl42- - H2 O - Pt H2 O Cl3- - Cl-At 25-C- it is found -d-PtCl42- -dt - 3-9-10-5 s-1-PtCl42- -2-1-10-3 Lmol-1 s-1 -PtH2 O Cl3- -Cl- - Value of KC when fourth Cl- is complexed in the reaction is-

For 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^/ of the required reaction =1/K_ ...

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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}

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1.86 \times 2

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53.85

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8.19 \times 10^{- 8}

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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 $= \frac{1}{K_{c}} = 53.85$

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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. Considering the elements

F, C l, O a n d S,

the correct order of their electron affinity values is:

Q. The reaction

S O_{2} {C l}_{2} \to S O_{2} + {C l}_{2}

, is a first order gas reaction with

k = 2.2 \times 10^{- 5} s^{- 1}

573

K

. What percent of

S O_{2} {C l}_{2}

is decomposed on heating at

573

K

for

90

m i n

Q. The correct order of the

1^{s t}

electron affinities of

N, O, F, S

and

C l

is:

Q. The table given below gives kinetic data for the following reaction at

298

K

O {C l}^{-} + I^{-} ⟶ O I^{-} + {C l}^{-}

Expt. No.	$[O {C l}^{-}]$ $(i n m o l$ ${d m}^{- 3})$	$[I^{-}]$ $(i n m o l$ ${d m}^{- 3})$	$[O H^{-}]$ $(i n m o l$ ${d m}^{- 3})$	$d [{I O}^{-}] / d t$ $(i n 10^{- 4} m o l$ ${d m}^{- 3}$ $s^{- 1})$
$1.$	$0.0017$	$0.0017$	$1.0$	$1.75$
$2.$	$0.0034$	$0.0017$	$1.0$	$3.50$
$3.$	$0.0017$	$0.0034$	$1.0$	$3.50$
$4.$	$0.0017$	$0.0017$	$0.5$	$3.50$

What is the rate law and what is the value of rate constant?

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The correct option is C 53.85For the given reaction, kf=3.9×10−5 s−1 kb=2.1×10−3 Lmol−1s−1 ∴Kc=kfkb=3.9×10−52.1×10−3 molL−1=0.0186 ∴K/c of the required reaction =1Kc=53.85