Question

question 5
plotting the inverse of the reactant concentration against time yields a straight line for:
a zero - order reaction
a first - order reaction
a second - order reaction
all chemical reactions

Explanation:

For a second - order reaction, the integrated rate law is $\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}$, where $[A]_t$ is the concentration of the reactant at time $t$, $k$ is the rate constant and $[A]_0$ is the initial concentration of the reactant. This equation is in the form of a straight - line equation $y = mx + c$ ($y=\frac{1}{[A]_t}$, $x = t$, $m = k$ and $c=\frac{1}{[A]_0}$). For a zero - order reaction, $[A]_t=-kt + [A]_0$ and for a first - order reaction, $\ln[A]_t=-kt+\ln[A]_0$. They do not have $\frac{1}{[A]_t}$ vs $t$ as a straight - line relationship. Not all chemical reactions have this relationship.

Answer:

C. A second - order reaction