1. Reaction Rate

2. Expressing rate of a chemical reaction:
Change in concentration of species/substance over time.
ex: 2N2O5(g) --> 4NO2(g) + O2(g)
Rate can be described by the increase in concentration of oxygen as
or by the decrease in [N2O5] as
Important note:
[O2] should be read as “concentration of O2” or “molarity of O2”

3. Looking at the coefficients, two molecules of N2O5 must decompose for each molecule of oxygen to form. So if we want to compare/relate rate of oxygen formation to the rate of N2O5 decomposition, we must take this into account. Also notice the negative sign on the right. Why is this? Ans: because N2O5 is being used up, so its change in concentration is negative.

4. Try this one:
N2 + 3H2 --> 2NH3 Write the equation that relates rate of production of NH3 to the rate at which hydrogen is used up.

5. Rate Laws
Equate rate of a reaction to concentrations of some reactant(s). ex: 2H2(g) + 2NO(g) --> N2(g) + 2H2O(g) Let’s say that it is found (experimentally) that, if we hold [NO] constant, doubling [H2] doubles R (rate). Therefore: rate (R) is proportional to [H2] or R = k[H2] (where k is some constant number) (We’ve just written a rate law.) Let’s say that it is also found that when [H2] is constant and [NO] is doubled, R increases fourfold, and when [NO] is tripled, R increases ninefold. Therefore: rate is proportional to the square of [NO] or R = k[NO]2 (another rate law) If we put the two together, we get: R = k[H2][NO]2 Therefore we can say that this reaction is 1st order with respect to H2. The reaction is 2nd order with respect to NO. The reaction is 3rd order overall. (just adding up the superscripts n the rate law)

6. For a single-step reaction, rate laws come from the coefficients in the balanced equation. ex: The rate law for: A + B --> 2C is R = k[A][B] This reaction requires one particle of each reactant to collide. Therefore doubling the concentration of either reactant will double the rate of reaction. But the reverse: 2C --> A + B gives the rate law R = k[C]2 because 2 molecules of C are required to decompose to create one A and one B. Therefore the rate is proportional to [C] x [C] or [C]2

7. Many reactions are the result of a sequence of steps
Many reactions are the result of a sequence of steps. For example, the following reaction NO2 + CO  NO + CO2 (overall reaction) Is actually believed to be the result of two steps: step 1: NO2 + NO2 --> NO3 + NO (this is slower) step 2: NO3 + CO --> NO2 + CO2 (this step is faster) Rate is determined by the slower step. (the "rate-determining step") Therefore the entire rate law for the overall reaction is R = k[NO2]2 If we thought that the overall reaction were single-step we would expect the rate law to be R = k[NO2][CO] but experimental data shows that this is not the case. [CO] is not included in the rate law because is a part of the faster step. Increasing the fast step does not speed the reaction because the reaction rate is limited by the slower step.