9 posts • Page 1 of 1
A first order rate law forms a linear relationship when the natural log of [A] is plotted over time. A second d order rate law forms this relationship when 1/[A] is graphed over time. We derive these relationships from the integrated rate laws of each order of reaction. Since a 0th order rxn doesn't depend on the concentration, the relationship of simply [A] over time gives a linear graph
All of the graphs will be linear, but the slopes will vary. For instance, the graph of a first-order and zero-order reaction will have a decreasing slope of -k while a second-order reaction will have an increasing slope of k.
For zero order, the rate law makes a linear graph, which means that the rate just goes at constant k (slope). For first order reactions, the graph is also linear if you plot ln[A] versus time. For second order reactions, the graph is linear is you plot 1/[A] versus time.
Depending on the different rate laws for the different ordered reactions, plotting the corresponding term on the y-axis against time will yield a straight line instead of a curve. For example, if given a graph with the y-axis as 1/ln[A] plotted against time, and the graph is a straight line, then you know it is a second order reaction. The same goes for a first and zero order reactions with y-axis ln[A] and [A], respectively.
The graphs for the different rate laws will all be linear, but what is plotted on the y axis varies. For a 0th order reaction, [A] vs time is linear, and the slop is equal to -k. For a 1st order reaction, ln[A] vs time is linear and the slope is equal to -k. For a 2nd order reaction, 1/[A] vs time will be linear and the slope is equal to k. The differences are important because looking a the graph can help us determine the order of the rate law for a given reaction.
Who is online
Users browsing this forum: No registered users and 3 guests