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JEE Advance - Chemistry (1996 - No. 14)

The ionisation constant of $$NH_4^+$$ in water is 5.6 $$\times$$ 10-10 at 25oC. The rate constant for the reaction of $$NH_4^+$$ and $$OH^-$$ to form NH3 and H2O at 25oC is 3.4 $$\times$$ 1010 L mol-1s-1. Calculate the rate constant for proton transfer from water to NH3.
6.07 × 103 L mol-1s-1
6.07 × 104 L mol-1s-1
6.07 × 105 L mol-1s-1
6.07 × 106 L mol-1s-1
6.07 × 107 L mol-1s-1

Explanation

Given, NH$$_4^ + $$ + H2O $$\rightleftharpoons$$ NH3 + H3O+

$${K_a} = {{[N{H_3}][{H_3}{O^ + }]} \over {[NH_4^ + ][{H_2}O]}}$$ (Ka = 5.6 $$\times$$ 10$$-$$10) ..... [1]

$$NH_4^ + + \mathop O\limits^ - H$$ $$\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{k_b}}^{{k_f}}} $$ $$N{H_3} + {H_2}O$$ (kf = 3.4 $$\times$$ 1010 Lmol$$-$$1 s$$-$$1) ...... [2]

$${K_{eq}} = {{{k_f}} \over {{k_b}}} = {{[N{H_3}][{H_3}O]} \over {[NH_4^ + ][O{H^ - }]}}$$

$$ = {{[N{H_3}][{H_3}{O^ + }]} \over {[NH_4^ + ][{H_2}O]}} \times {{{{[{H_2}O]}^2}} \over {[{H_3}{O^ + }][O{H^ - }]}} = {{{K_a}} \over {{K_w}}}$$

$$\therefore$$ $${k_b} = {{{K_w}} \over {{K_a}}} \times {k_f} = {{1.0 \times {{10}^{ - 4}}} \over {5.6 \times {{10}^{ - 10}}}} \times 3.4 \times {10^{10}} = 6.07 \times {10^5}$$

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