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

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The ionisation constant of N H _ 4 ^ + in water is 5.6 × 1 0 ^ - 10...

The ionisation constant of NH4+N{H}_{4}^{+} in water is 5.6 ×10−10\times 1{0}^{- 10} at 25 °C° C. The rate constant for the reaction of NH4+N{H}_{4}^{+} and OH−OH^{-} to form NH3N{H}_{3} and H2O{H}_{2}O at 25°C25 ° C is 3.4×1010/mol−1S−13.4 \times 1{0}^{10} / mol^{- 1}S^{- 1}. The rate constant for proton transfer from water to NH3N{H}_{3} is

Options

  1. A6.07×1056.07 \times 1{0}^{5}
  2. B0.607×1050.607 \times 1{0}^{5}
  3. C60.7×10560.7 \times 1{0}^{5}
  4. D6.07×10106.07 \times 1{0}^{10}

Correct answer

A. 6.07×1056.07 \times 1{0}^{5}

Step-by-step solution

NH4+⇌NH3+H+;k1=5.6×1010N{H}_{4}^{+} ⇌ N{H}_{3} + H^{+}; {k}_{1} = 5.6 \times 1{0}^{10} …(i)
H2O⇌H++OH;k2=1×1014{H}_{2}O ⇌ H^{+} + OH; {k}_{2} = 1 \times 1{0}^{14} …(ii)
K=k1k2=5.6×10−101×10−14K = \dfrac{{k}_{1}}{{k}_{2}} = \dfrac{5.6 \times 1{0}^{- 10}}{1 \times 1{0}^{- 14}}
=5.6×104= 5.6 \times 1{0}^{4}
By subtracting equation (ii) from equation (i) we get
NH4++OH⇌NH3+H2ON{H}_{4}^{+} + OH ⇌ N{H}_{3} + {H}_{2}O
From equation
kf=3.4×1010/mil−1S−1{k}_{f} = 3.4 \times 1{0}^{10} / mil^{- 1}S^{- 1}
NH3+H2O⇌NH4++OH−N{H}_{3} + {H}_{2}O ⇌ N{H}_{4}^{+} + OH^{-}
kr=?{k}_{r} = ?
K=kfkrK = \dfrac{{k}_{f}}{{k}_{r}}
So kr=kfK=3.4×10105.6×104{k}_{r} = \dfrac{{k}_{f}}{K} = \dfrac{3.4 \times 1{0}^{10}}{5.6 \times 1{0}^{4}}
=6.07×105= 6.07 \times 1{0}^{5}
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