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

Correct answer: 6 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

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The ionisation constant for N H _ 4 ^ + in water is 5.0 × 1 0 ^ -...

The ionisation constant for

N{H}_{4}^{+}

in water is

5.0 \times 1{0}^{- 10}

{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

{25}^{o}C

3.0 \times {10}^{10} {\text{Lmol}}^{- 1} {\text{s}}^{- 1}

. The rate constant for proton transfer from water to

N{H}_{3}

x \times 1{0}^{5}

. The value of 'x' is 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

Correct answer

Step-by-step 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,

{\text{K}}_{\text{c}} \text{=}\dfrac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{b}}}

5 \times {10}^{4} = \dfrac{3 \times {10}^{10}}{{\text{k}}_{\text{b}}}

{\text{k}}_{\text{b}} = 6 \times {10}^{5}

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