# Chapter 14 Chemical Kinetics: The Rates of Chemical Reactions - Study Questions - Page 553f: 58

#### Work Step by Step

a) For a first-order reaction: $\ln(C/C_0)=-kt;k=0.693/t_{1/2}$, at constant volume, we can use this equation in mass basis: $\ln(m/1.50\ mg)=-0.693/3.9\ h\dot{}5.25\ h$ $m=0.59\ mg$ b) $\ln(2.5\dot{}10^{-6}\ mg/1.50\ mg)=-0.693/3.9\ h\dot{}t$ $t=74.87\ h$

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