And that's how we find our K.
Now, in other words, to summarize, if after graphing our concentration of reactants versus time, we get a negative slope like this, we get a graph like this. That means our A must be one. So now let's look at what the graph would look like if our A was actually one, if our order was first order or A equals one. Now once again we begin the same way. Negative change in concentration of x over change in time is equal to our constant KDF times the concentration of x to the one power. So now this guy remains.