How to calculate the rate constant of a zero order reaction? The reactant concentration in a zero-order reaction was $7.00×10^{−2}~\mathrm{M}$ after $160~\mathrm{s}$ and $2.50×10^{−2}~\mathrm{M}$ after $345~\mathrm{s}$. What is the rate constant for this reaction?  
  What was the initial reactant concentration for the reaction described in part one?   
  The reactant concentration in a first-order reaction was $7.70×10^{−2}~\mathrm{M}$ after $50.0~\mathrm{s}$ and $1.60×10^{−3}~\mathrm{M}$ after $85.0~\mathrm{s}$. What is the rate constant for this reaction? 
  


Background info: The integrated rate laws for zero and first order reactions may be arranged such that they resemble the equation for a straight line, 
$y=mx+b$. 

\begin{array}{lccc}
\text{Order}&amp; \text{Integrated Rate Law}&amp; \text{Graph}&amp; \text{Slope} \\\hline
  0 &amp;  [A]=−kt+[A]_0      &amp; [A]\text{ vs. }t    &amp; −k\\ 
  1 &amp; \ln[A]=−kt+\ln[A]_0 &amp; \ln[A]\text{ vs. }t &amp; −k\\\hline
\end{array}
According to your table, for a zero order reaction, you make a graph of the measured concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.

For a first order reaction, you make a graph of the natural log of the concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.