Integrated Rate Laws

Quick Review Our previous lessons have focused on calculating the rate of a reaction and rate constant by observing how changes in concentration affect the initial rate of a reaction. ExperimentInitial [A]Initial [B]Initial Rate of the Reaction 11.0 M 2.0 M/sec 21.0 M2.0 M4.0 M/sec 32.0 M 16.0 M/sec

New Stuff What we now focus on is looking at the concentration of a reactant over time. We are also going to focus on graphs of these reactions, and that will help us decide what order of reaction and what equation to use when solving for the reaction constant (k). [A] (M)Time (Sec) 2.0 M0.0 sec 1.8 M1.5 sec 1.6 M3.0 sec

Integrated Rate Laws; The Good and Bad Good News: When we look at these types of problems, they will only be 0 th, 1 st, or 2 nd order reactions. That’s it! Bad News: There’s a different equation for each of them. Good News: I’ll give you the equations on the test. We will look at graphs to determine which order reaction. Bad News: All the graphs are really similar so you have to be paying attention.

Y = mx + b We are going to be using graphs and data to solve for the reaction constant (k). We will be arranging our equations into the general format y = mx + b. What is special about this?? In all equations the m (slope of the line) will equal k (reaction constant) M = k for our equations. Solve for the slope, and you will be solving for k.

The quest for a straight line… What is the slope of this line? This question is difficult because depending on where we take our measurements, the slope changes!

Our goal is to solve for the reaction constant. What does the word constant imply? We will start with graphs of concentration of a compound changing over time. If the reaction constant = slope, then we MUST have a graph that shows a constant slope. If that graph is not given, we must manipulate the graphs until we have a SRAIGHT LINE!

0 th Order Reactions What does it mean if a reaction is 0 th order? That means that the rate of the reaction is independent of the concentrations of the reactants. It doesn’t matter how much of the compound you add, the rate of the reaction does not change. For the equation A  C, the rate law will look like this for 0 th order: Rate = k

0 th Order Graph If we graph Concentration vs. Time, what will it look like if the reaction is 0 th Order? Is the slope a constant? It is! If the slope of a concentration vs. time graph is constant, it indicates a 0 th order reaction.

0 th Order Equation If we wanted to calculate the rate constant for a 0 th order reaction, we use this specific equation: [A]= -kt + [A] 0 [A] = Concentration at a specific time K = rate constant T = time (check for units) [A] 0 = Initial concentration, Concentration at = 0.

1 st Order Reactions 1 st reactions are the most common that occur. What does it mean if a reaction is first order? Rate and Concentration are related by a power of 1. Meaning that if we double the concentration of [A], we will double the rate of the reaction. Tripling the concentrations will triple the rate of the reaction, ect. The rate law might look like this: A  C for 1 st Order Reaction, Rate = k[A]

1 st Order Graph If we graph Concentration vs. Time, what will it look like if the reaction is 1 st Order ? Is the slope a constant? NO IT IS NOT! So remember, we can only do our calculations when the slope is constant, so how do we get a constant slope?

1 st Order Graph If our relationship is 1 st Order, we can take the derivative of the line to “straighten” it out. Without getting too complicated, taking the derivative allows us to look at the linear relationship between the X and Y axis. Is the slope a constant? It is! If the slope of ln[A] vs. time graph is constant, it indicates a 1 st order reaction.

1 st Order Equation If we wanted to calculate the rate constant for a 1 st Order reaction, we use this specific equation: ln[A]= -kt + ln[A] 0 ln = Natural Log – It’s just a button on the calculator. [A] = Concentration at a specific time K = rate constant T = time (check for units) [A] 0 = Initial concentration, Concentration at t = 0.

2 nd Order Reactions What does it mean if a reaction is 2 nd order? Rate and concentration are related by a power of 2. [A] 2 = Rate of the reaction That means that if we double the concentration, the rate quadruples….. [2] 2 = 4 [3] 2 = 9 [4] 2 = 16 For the equation A  C, the rate law looks like this for 2 nd order: Rate = k[A] 2

2 nd Order Graph If we graph Concentration vs. Time, what will it look like a 2 nd Order reaction? Is the slope a constant? NO IT IS NOT! So remember, we can only do our calculations when the slope is constant, so how do we get a constant slope?

2 nd Order Graph BAM! I GOT THIS. TAKE THE NATURAL LOG AND WATCH THAT LINE STRAIGHTEN OUT! Is the slope a constant? NO IT IS NOT! Wait a minute….I though taking the natural log (aka taking the derivative) made straight lines?!?! It does, but only for what order reactions?

2 nd Order Graph So how do I straighten out a 2 nd order reaction? You must take the inverse or 1/[A]. Is the slope a constant? Finally! It is!! If the slope of 1/[A] vs. time graph is constant, it indicates a 2 nd order reaction.

2 nd Order Equation If we wanted to calculate the rate constant for a 2 nd order reaction, we use this specific equation: 1/[A]= kt + 1/[A] 0 *Notice in this equation that k is positive [A] = Concentration at a specific time K = rate constant T = time (check for units) [A] 0 = Initial concentration, Concentration at t = 0.

Quick Review of Things to Look For: Remember it’s all about finding the straight lines. The straight line means the slope is constant and remember, slope = rate constant! If the [A] vs. time graph is straight, then it’s a 0 th order reaction. If the ln [A] vs. time graph is straight, then it’s a 1 st order reaction. If the 1/[A] vs. time graph is straight, then it’s a 2 nd order reaction.

PRACTICE It’s time for your favorite game show, “WHAT ORDER REACTION AM I BASED ON THE GRAPHS!!!!!!!!!!!!!!!!!!”

Here’s the first contestant!

Bachelorette #2

What’s Behind Curtain #3!

Graphing Worksheet Practice