A generic reaction of aA———> products, gives a straight line plot of ln[A] versus time with a slope of -7.71x10^-3 Ms^-1. Assuming that the initial concentration of A is 0.135M, calculate the time in second,t, required for 25.6% of the original sample to react.
The given generic reaction is a First Order reaction.
Slope = -k = "-7.71\\times 10^{-3}"
Hence, k = "7.71\\times 10^{-3}"
Initial concentration = a = 0.135M
We know in the first order reaction,
"t = \\dfrac{2.303}{k}log\\dfrac{a}{a-x}"
"t= \\dfrac{2.303}{7.71\\times 10^{-3}}log\\dfrac{0.135}{0.135 - \\dfrac{25.6}{100}\\times 0.135}"
Hence, "t = 38.35" s
Comments
Leave a comment