If the equilibirium reaction is
2 03(g) <--> 3 O2 9(g) is first order with respect to ozone, what type of plot( meaning what is plotted on the y axis vs what is plotted on the x-axis) will yield a straight line? what will the slope of the line represent?
PLEASE help im stumped and have a test.
To determine what type of plot will yield a straight line and what the slope of the line represents for a first-order reaction, we need to understand the characteristics of a first-order reaction.
A first-order reaction is a reaction whose rate depends directly on the concentration of one reactant. In your case, the reaction is first order with respect to ozone (O3).
To determine the order of the reaction, we can use the integrated rate law equation for a first-order reaction:
ln([A]t /[A]0) = -kt
where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is time.
By rearranging the equation, we can obtain a linear relationship:
ln([A]t) = -kt + ln([A]0)
Now, we can see that plotting ln([A]t) on the y-axis and time (t) on the x-axis will yield a straight line for a first-order reaction.
The slope of this line represents the rate constant (k) of the reaction. The negative value of the slope is equal to the rate constant multiplied by -1. Therefore, by measuring the slope of the line, you can determine the rate constant for the reaction.
To summarize:
- Plot ln([A]t) on the y-axis
- Plot time (t) on the x-axis
- The slope of the line represents the rate constant (k) of the reaction.