Question

For a zero ordue reaction, ( t^{prime} 2=left[begin{array}{ll}A]_{0} / 2 k & begin{array}{l} text { Al } ]_{0} text { -initial conc. } k=text { rate constant }end{array}end{array}right. )
Rate law ( left.Rightarrow[A]_{t}=-k t+[A]right}[A]_{t}=cos _{t i m e^{i} t^{prime}} )
( S_{D} )
( [A]_{t}=-k t+[A]_{0} )
( A_{t}=0.08 mathrm{M} 0.08=-mathrm{K} times 10+0.1 )
( mathrm{A}_{0}=0.1 mathrm{M} quad 0.08-0.1=-mathrm{K} times 10 )
( t_{1}=mathrm{lomin} quad ? )
Now, ( t Z_{2}=frac{[A]_{0}}{2 K}=frac{0.1}{2 times 2 times 10^{-3}} )
( therefore t^{prime} / 2=25 mathrm{min} )
Therefolc time taken to complete the reaction will be 2 hall-linus
2 hafferes ( =2 times 25 ) min ( =50 ) min

# For the zero order reaction A-B-C: initial concentration of A is 0.1M. If [A] = 0.08 M after 10 minutes, then its half-life and time for completion are respectively: A 10 min; 20 min B 2x10-min, 4x10-3 min C 25 min; 50 min D 250 min, 500 min

Solution