$y=mx+b[A]=-kt+[A{]}_{0}At-kln[A]=-kt+ln[A{]}_{0}ln[A]t-k\frac{1}{[(A)]}=kt+\frac{1}{[(A){]}_0}\frac{1}{[(A)]}tk+-$ Using Integrated Rate Laws

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

$\begin{array}{cc}\text{O}\text{r}\text{d}\text{e}\text{r}& \text{I}\text{n}\text{t}\text{e}\text{g}\text{r}\text{a}\text{t}\text{e}\text{d}\text{R}\text{a}\text{t}\text{e}\text{L}\text{a}\text{w}\\ \text{G}\text{r}\text{a}\text{p}\text{h}& \text{S}\text{l}\text{o}\text{p}\text{e}\\ 0& A\\ Avs.t& -k\\ 1& ln[A]\\ ln[A]vs.t& -k\\ 2& \frac{1}{[(A)]}=kt+\frac{1}{[(A){]}_0}\\ \frac{1}{[(A)]}vs.t& k\end{array}=kt+\frac{1}{[(A){]}_0}=-kt+ln[A{]}_0=kt+\frac{1}{[(A){]}_0}=-kt+ln[A{]}_0=-kt+[A{]}_0$

Part A

The reactant concentration in a zero$-$order reaction was $7\mathrm{.}00{10}^{-2}M$ after $140s$ and $1\mathrm{.}50{10}^{-2}M$ after $330s.$ What is the rate constant for this reaction?

Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

View Available Hint$($s$)$

$k_{0th}=$