The ionisation constant of N H 4 + in water is 5.6 × 1 0 - 10 at 25 ° 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 ° C is 3.4 × 1 0 10 / m o l - 1 S - 1 . The rate constant for proton transfer from water to N H 3 is

Step-by-step solution

N H 4 + ⇌ N H 3 + H + ; k 1 = 5.6 × 1 0 10 …(i) H 2 O ⇌ H + + O H ; k 2 = 1 × 1 0 14 …(ii) K = k 1 k 2 = 5.6 × 1 0 - 10 1 × 1 0 - 14 = 5.6 × 1 0 4 By subtracting equation (ii) from equation (i) we get N H 4 + + O H ⇌ N H 3 + H 2 O From equation k f = 3.4 × 1 0 10 / m i l - 1 S - 1 N H 3 + H 2 O ⇌ N H 4 + + O H - k r = ? K = k f k r So k r = k f K = 3.4 × 1 0 10 5.6 × 1 0 4 = 6.07 × 1 0 5