If you thought all of these factors are used in the Arrhenius equation, consider that the rate constant is independent of the reactant concentrations. The concentration effect is considered in the rate law by multiplying the concentrations by the constant, but the concentrations themselves are independent of the actual Arrhenius calculation.

The calculation shows that . In this equation, is the activation energy and is the temperature.

Keep in mind that the chemical equation and its coefficients have NOTHING to do with the rate law. The order of each reactant must be determined by experiment.

To find the order of each reactant, compare the initial reaction rates of two trials in which only one of the three reactants' concentrations is altered. For example, trials 1 and 4 keep A and B equal, but C is doubled. When C is doubled, we see that the initial reaction rate is quadrupled. As a result, we determine that the order of reactant C is 2. When the reactant is altered, but the initial reaction rate is kept constant, as seen in trials 1 and 3 with respect to A, the order of that reactant is 0. Finally, when the reactant is multiplied by the same factor that the initial reaction rate is multiplied, as seen in trials 1 and 2 with respect to B, the order of the reactant is 1.

Putting the data together: A is zeroth order, B is first order, and C is second order. Our rate law can thus be written .

The reaction table in the passage indicates that the reaction rate varies in a 1-to-1 fashion as you vary the each reactant, while holding the other constant.

The rate law is written as .

Compare trials 1 and 2 to see that doubling the ammonium concentration doubles the rate. The reaction is first order for ammonium: .

Compare trials 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate: .

The final rate law is .

For the reaction given:

We need to find the reaction order, therefore we have to find the values of a and b in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of Y was doubled in run 2 while keeping the concentration of X constant. We can observe that the reaction rate was quadrupled. Let's compare by ratios:

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

Plugging the values into the equation gives:

Simplifying the above equation gives:

Therefore, 

We can also compare the results of run number 2 and run number 3. In that case, the concentration of Y was kept constant while the concentration of X was doubled. We can see that the reaction rate doubled. Therefore the reaction is first order with respect to X and second order with respect to Y. Therefore, the reaction is third order overall:

For the reaction given:

We need to find the reaction order, therefore we have to find the values of x and y in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of A was doubled in run 2 while keeping the concentration of B constant. We can observe that the reaction rate was doubled. Let's compare by ratios:

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

Plugging the values into the equation gives:

Simplifying the above equation gives:

Therefore, 

We can also compare the results of run number 2 and run number 3. In that case, the concentration of A was kept constant while the concentration of B was decreased by half. We can see that the reaction rate decreased by half. Therefore the reaction is first order with respect to A and first order with respect to B. Therefore, the reaction is second order overall:

Point 3 is where you would expect to find a relatively stable intermediate. An intermediate is more stable than a transition state, but not as stable as the original reactants and final products. Stability is inversely proportional to energy, thus we are looking for the point that is between the highest and lowest energies in the reaction. By this logic, point 1 is the reactants, 2 and 4 are transition states, 3 is a stable intermediate, and 5 is the products.

Most organic reactions are carried out in multiple steps. The rate equation can be derived from the process that occurs in the slowest step of the mechanism. An E1 reaction's slow step is when a leaving group separates from the hydrocarbon and a carbocation is formed. The only reactant in this step is one molar equivalent of the hydrocarbon. Thus, the rate equation only depends on that substance. The molar equivalent determines the exponent of each reactant in the equation.