The Integrated Rate Laws are used to determine the amount of reactants or products as a function of time. There will be times where *the* *order of the reaction isn’t directly given to us* within a question. In these difficult times just remember these 4 helpful tips in determining the order of the reaction.

**Tip #1 Radioactivity and Order**

Some tips will require a little math, or processing plots and/or key phrases, but luckily the first tip is pretty easy to spot.

**If the question uses the word “radioactive” or gives an element with its atomic mass like Carbon-13 then the reaction is 1 ^{st} order.**

So let’s take a look at our first example and keep an eye out for the word “radioactive” or for an element with its atomic mass.

**EXAMPLE:** Plutonium-239 is used in nuclear reactors of some power plants and has a rate constant equal to 2.84 x 10^{-6} yr ^{–1}. If there are 4.00 x 10^{2} g of isotope in a small reactor, how long will it take for the substance to decay to 1.00 x 10^{2} g?

*HINT: **We are dealing with a first order process because we are given an element and its atomic mass in the form of Plutonium-239. In order to solve the question we follow the steps given below: *

**STEP 1:** Isolate the given variables from the question.

**STEP 2: **Utilize the Integrated Rate Law for the chemical reaction.

**Tip #2 Plot of a Graph**

The plot of a graph is of the y-axis versus the x-axis. By applying the equation of a straight line to an Integrated Rate Law equation we can determine the order.

**a. If the question states that a plot of “[reactant] vs. time gives a straight line” then it’s a zeroth order reaction.**

**EXAMPLE:** A student studied the kinetics of the reaction: C_{2}H_{4 }(g) → C_{2}H_{2 }(g) + H_{2 }(g) by monitoring the C_{2}H_{4 }concentration (in moles/L) as a function of time. The student made a plot of* *[C_{2}H_{4}] against t. The plot was a straight line with the equation y = −0.618x − 1.427 (where y =* ln *[C_{2}H_{4}] and x = t). What is the order of the reaction?

*SOLUTION: **Since the question states, “*The student made a plot of* *[C_{2}H_{4}] against t*”, we are dealing with a zeroth order process. As was previously mentioned a zeroth order process gives a plot of *

**[reactant] vs. time**

*as a straight line. In this case, the [reactant] is represented as [C*

_{2}H_{4}].**b. If the question states that a plot of “ln[reactant] vs. time gives a straight line” then it’s a first order reaction.**

**EXAMPLE:** For the reaction of Cyclopropane (g) → propene(g) at 500°C, a plot of ln [cyclopropane] vs. t gives a straight line with a slope of -0.00067 s^{-1}. What is the order of this reaction?

*SOLUTION: **We are dealing with a first order process because we are given a plot of *

**ln[reactant] vs. time**

*that gives a straight line. In this case, the ln[reactant] is represented as ln[cyclopropane].*

**c. If the question states that a plot of “1/[reactant] vs. time gives a straight line” then it’s a second order reaction.**

**EXAMPLE:** A reaction originally has 0.37 M of a reactant and after 75 seconds there are 0.15 M of the same reactant present. A plot of the inverse of the reactant concentration versus time results in a straight-line. What is the order for this reaction?

*SOLUTION: **The inverse of the reactant concentration means 1/[reactant]. We are dealing with a second order process because we are given a plot of *

**1/[reactant] vs. time**

*that gives a straight line.*

Now let’s apply all we’ve learned at this point to determine the answer to the following example question.

**EXAMPLE: **For the reaction 2 NO_{2} (g) → 2 NO (g) + O_{2 }(g), a graph of ln[NO_{2}] versus time is a straight line with a slope of – 0.842 s^{–1} at 50 °C. Determine the rate law and rate constant of the reaction.

*HINT:** We are dealing with a first order process because we are given a graph of *

**ln[NO**

_{2}] versus time*that gives a straight line. In order to solve the question we follow the steps given below:*

**STEP 1:** Identify the given variables from the question.

*HINT: **Since we’ve become experts at spotting the order of a plot or graph this is going to be easy. We know that ln[reactant] vs. time represents a first order process. Using this fact will allow us to determine the rate law and rate constant of a first order process. *

**STEP 2**: Isolate the given variables and solve.

As we mentioned earlier, since the graph is a plot of **ln[reactant] versus time** it must be a **first order** process.

The slope of a straight line is represented by the variable **m**, which also represents the rate constant, **k**.

The rate law is based on the order of the reaction. Since it’s a **first order** reaction the reaction order would be equal to **1**. So the rate law would be given as:

**Tip #3 Initial Concentration and Half-Life**

**If the question states, that “the half-life of a reaction is independent of the initial concentration of the reactant” then the reaction is 1^{st} order.**

Presented below are the half-life equations for a 0^{th} order, 1^{st} order and 2^{nd} order reaction.

Notice how the initial concentration, [A]_{0}, of a reactant is found in both the equations for a 0^{th} order and 2^{nd} order reaction, but not for a 1^{st} order reaction. This illustrates how the half-life for a 1^{st} order reaction is independent of the initial concentration because of its absence in the equation.

**EXAMPLE: **The half-life for the radioactive decay of radium is 1620 years while remaining independent of the initial concentration of the reactant. What is the initial concentration if it took 6000 years for the final concentration to reach 0.110 M while the rate constant is 4.28 x 10^{–4 }yr^{ –1}?

** ****STEP 1:** Identify the given variables from the question.

**STEP 2:** Determine the order and the integrated rate law equation needed.

Since the question states **half-life is independent of initial concentration** it must be a **first order** process and its integrated rate law equation is given as:

**STEP 3: **Plug in the given variables into the Integrated Rate Law Equation for the first order process.

Solving for ln[0.110] and multiplying the rate constant, k, and time, t, gives:

Adding 2.568 to both sides of the equation gives:

To isolate the initial concentration, [A]_{0}, we take the inverse of the natural log.

**Tip #4 Rate Constant Units and Order **

To understand the usefulness of Tip #4 let’s take a look at the following example.

**EXAMPLE: **A reaction has a rate constant of 0.0185 s^{-1} and the only reactant has an initial concentration of 0.135 M. What is the concentration in molarity of this reactant after 125 seconds?

*HINT: **In this particular example we aren’t given the order of the reaction and if we refer to our previous three tips none of them will help. The key to determining the order of the reaction will involve looking at the units for the rate constant, k.*

**If the question gives the units for rate constant, k, we can predict the order of the reaction by use the following equation where n represents the order of the reaction.**

**STEP 1: **Look at the units for the rate constant, k, and make an assumption about the possible order.

The rate constant is given as 0.0185 s^{-1}. This means that the molarity (M) has been removed and that is only possible if **n** = 1.

**STEP 2: **After your assumption, take your value for **n** and see if you obtain the desired units for the rate constant, k.

Based on our assumption we obtain:

Remember that any variable with a power of zero will be equal to 1.

In the end we obtain the same units for the **rate constant** from the question.

*SOLUTION: **We have just proven that n = 1 and can conclude we have a first order process since n equals the order of the reaction. This means that in order to determine our final concentration we can utilize the Integrated Rate Law for a first order process.*

Let’s try a few more examples utilizing the equation to find the units for the rate constant, k.

**EXAMPLE:** If we were told that the rate constant is equal to 0.125, what would be the units of k for a zeroth order reaction?

*SOLUTION: **Since n = 0 when we plug it into the equation for the rate constant k we obtain molarity times the inverse seconds. *

**EXAMPLE:** How about for a first order reaction?

*SOLUTION: **In this example n = 1 and so the units for k reduce down to inverse seconds.*

**EXAMPLE:** Hopefully you have the hang of it by now so determine the units of k for a second order reaction and then scroll down to check.

*SOLUTION: **At this point you’re probably thinking, “Who needs this solution? I’ve mastered this topic.” With ***n** equal to 2 we obtain the inverses for both molarity and seconds as the units for k.

Now that we’ve gone over these 4 helpful tips go out and calculate reactant concentrations, rate constant, time or half-life.