The ionisation constant for N H 4 + in water is 5.0 × 1 0 - 10 at 25 o C .The rate constant for the reaction of N H 4 + and O H - to form N H 3 and H 2 O at 25 o C is 3.0 × 10 10 Lmol − 1 s − 1 . The rate constant for proton transfer from water to N H 3 is x × 1 0 5 . The value of 'x' is

N H 4 + ⇌ N H 3 + H + K = 5 × 1 0 - 10 Now, H + + O H - ⇌ H 2 O K w = 1 0 14 now, K c = k f k b 5 × 10 4 = 3 × 10 10 k b k b = 6 × 10 5

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ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

The ionisation constant for $N H_{4}^{+}$ in water is $5.0 \times 1 0^{- 10}$ at $25^{o} C$ .The rate constant for the reaction of $N H_{4}^{+}$ and $O H^{-}$ to form $N H_{3}$ and $H_{2} O$ at $25^{o} C$ is $3.0 \times 10^{10} {Lmol}^{- 1} s^{- 1}$ . The rate constant for proton transfer from water to $N H_{3}$ is $x \times 1 0^{5}$ . The value of 'x' is

Answer:

6.00

Solution:

$N H_{4}^{+} ⇌ N H_{3} + H^{+} K = 5 \times 1 0^{- 10}$
Now,
$H^{+} + O H^{-} ⇌ H_{2} O K_{w} = 1 0^{14}$

now,
$K_{c} = \frac{k_{f}}{k_{b}}$
$5 \times 10^{4} = \frac{3 \times 10^{10}}{k_{b}}$
$k_{b} = 6 \times 10^{5}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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