Answer:
Ater 3 minute
product = [tex]1.629\times 10^{-5}[/tex]Molar
susbtrate = 0.081%
after 5.3 minutes
product =[tex] 2.879\times 10^{-5} M[/tex]
susbtrate = 0.01435
Explanation:
Given data:
[tex]Km = 1.2\times 10^{-4} M[/tex]
Sustrate concentration = 0.02 M
Amount of product [tex]2.7\mu moles/ltr[/tex]
Rate = dP/dT = product formed per time
[tex]= \frac{2.7 \times 10^{-6} moles}{30} = 9 \times 10^{-8[/tex]
we know rate [tex]= \frac{v_{max}[S]}{Km + S}[/tex]
hence [tex]9\times 10^{-8} = \frac{v_{max}(0.02)}{(1.2\times 10^{-4} + 0.02)}[/tex]
[tex]v_{max} = 9.054\times 10^{-8}[/tex]
now after 3 min = 180 sec
product formed [tex]dP = rate \times dT = 9.054\times 10^{-8}\times 180 = 1.629\times 10^{-5}[/tex]Molar,
% substrate utilised[tex] = \frac{(1.629\times 10^{-5})}{0.02} \times 100 = 0.081[/tex]%
after 3.5 minutes = 318
[tex] dP = 9.054x10^{-8}\times 318 = 2.879\times 10^{-5} M[/tex]
% [tex]of susbtarte\ utilised =\frac{2.879\times 10^{-5}}{0.02} \times 100 = 0.0143[/tex] %
