Now the integrated rate law describes the relationship between reactant and their concentrations as well as time. This helps to determine how long it takes for X amount of moles per liter of reactant to become consumed or used up. And we're going to say here that the integrated rate law depends on the order of the reaction.

Now, the first one we talked about is our zero order integrated rate law. Here, this is for reactions following zero order rate laws and with it we're going to use the following equation: \(A_{\text{T}} = -Kt + A_{0}\) where \(A_{\text{T}}\) equals the final reacting concentration, \(A_{0}\) equals the initial reacting concentration (because zero time has passed), \(K\) equals your rate constant in. Now remember to figure out the units for \(K\), we say \(K = M^{2 \text{ to the negative n} + 1} \times \text{time}^{-1}\). And we're going to say that \(n\) equals the order of the reaction. So if we're dealing with zero order, that means it is equal to zero. So \(0^{-0} + 1\) just means \(+1\). This would mean that for a zero order reaction, the units for \(K\) are in molarity to the \(1 \times \text{time}^{-1}\). Now, time here could be seconds, days, years, or whatever. Here, \(T\) would be our time. So these are the components that make up our zero order integrated rate law equation.

Now this equation is also connected to the equation of a straight line here, \(A_{\text{T}}\) would be connected to \(y\), your rate constant, \(K\), would be connected to \(m\) which is your slope. And remember it's not just \(K\) it's negative \(K\). \(T\) here will be connected to \(x\), and then your initial concentration connected to \(b\). If we look at this graphically, remember a graph is a plot of \(y\) versus \(x\). Your \(y\) again would be your \(A_{\text{T}}\). So the concentration of your reactant and then \(T\) is your \(x\). And then remember here this is our initial concentration. When we first started our reaction notice that the slope is decreasing because remember slope is equal to \(m\), which is equal to negative \(K\). Negative \(K\) meaning that it's decreasing over time. Also, remember that slope is equal to change in \(y\) over change in \(x\), which in this case we can say is really the change in concentration over the change in time. So keep these in mind when discussing any question dealing with a zero order reaction.