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$64 M \overset{I}{⟶} 32 M \overset{II}{⟶} 16 M \overset{III}{⟶} 8 M \overset{IV}{⟶} 4 M$ $\overset{V}{⟶} 2 M \overset{VI}{⟶} 1 M .$
Steps I and II follow first order kinetics, III & IV follow zero order kinetics. V and VI steps follow second order kinetics. If the half lives for the reactions with initial concentrations 64 M, 16 M, 4 M, are 10 min, 20 min, 40 min respectively. Total time taken for all the steps.

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70 min

170 min

140 min

160 min

answer is B.

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Detailed Solution

For 1st order reaction

$t_{75 %} = 2 t_{\frac{1}{2}} = 2 \times 10 = 20 \min$

For zero order reaction

$t_{\frac{3}{4}} = 1.5 t_{\frac{1}{2}} = 1.5 \times 20 = 30 \min$

For second order reaction

$t_{\frac{3}{4}} = 3 t_{\frac{1}{2}} = 3 \times 40 = 120 \min$

Total time = 20+30+120 = 170 min

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