From my understanding plotting the conc. over time for a 0-order reaction yield a negative linear slope. Plotting conc vs time for a first order reaction intuitively yields an exponential type slope. however, my teacher never covered the looks of a conc vs time graph for a second order reaction. What may this look like. In conjunction, what does it look like when we graph its rate over concentration?


1 Answer 1


Let's take a single-component second-order reaction:

$$\ce{2A -> B}$$

If it is a 2nd-order reaction, the rate will be proportional to the concentration of $A$, squared:

$$\frac{dA}{dt} = -kA^2$$

That equation can be integrated:

$$\frac{dA}{A^2} == -k dt$$

$$-(A^{-1}-A_0^{-1}) = -k(t-t_0)$$

And solving for $A$:

$$A = \frac{1}{k(t-t_0) + A_0^{-1}}$$

That is the equation of a hyberbola. It goes to zero more slowly than a the expression for a first-order reaction, i.e. an exponential decay. This makes sense because as $A$ gets more and more diluted as the reaction occurs, collisions between $A$ molecules become rarer and rarer.

See this web page for more information.


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