Li B. answered • 09/30/15

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Experienced HS Teacher with Expertise in Math and Science

The integrated rate law for a first order reaction is:

[A] = [A

_{0}] e^{-kt }where [A] = concentration of A at time t and [A_{0}]= concentration of A at the very beginning (t=0)Here, we know that when t=5hr, [A] = 0.17 [A

_{0}] (17% of original)Substitute into equation:

0.17 [A

_{0}] = [A_{0}] e^{-k(5)}Divide both sides by [A

_{0}]0.17 = e

^{-5k}Take natural log of both sides

ln 0.17 = -5k

solve for k

k = - (ln 0.17)/5

Repeat for half life substituting original equation with [A] = 0.5 [A

_{0}] and t = T for the half life (half life is the time it takes for the concentration to reduce by half, or 50%)0.5 [A

_{0}] = [A_{0}] e^{-kT}Do the same thing (divide both sides by [A

_{0}] and take natural log as before:ln 0.5 = - kT

Substitute k from earlier

ln 0.5 = - (- (ln 0.17)/5)*T

Solve for T

-0.693 = -(-(-1.772)/5)*T

0.693 = 0.354 * T

T = 1.957 = 2 hours is the half-life

Sephi K.

Im confused wouldn’t it be 83% of the original not 17%07/21/20