Question

what are you going to do if there's an initial rate given but it's for ONLY ONE REACTANT? From my example, the given is 'Initial Rate = -d [NO]/dt (M/s). I thought it must be the initial rate FOR THE WHOLE EXPERIMENT? PLEASE HELP!

Answer 1

When you're talking about the rate of the reaction, you're actually ALWAYS talking about a specific compound, rather than the "whole reaction". Look at the way the reaction rate is being measured - in M/s, a change in concentration per unit time. A "reaction" doesn't have a concentration, but the specific reactants and products do, and these concentrations change over the course of the reaction as reactants are used up and products are formed.

In your example, NO is the reactant, and -d[NO]/dt is the initial rate of change. Notice it's negative - this means the concentration is DECREASING with time, which is what you'd expect for a reactant being used up in a reaction.

The only thing you need to be careful about is the stoichiometry of the reaction. If you have a simple reaction like AB --> A + B, each mole of AB produces a mole of A and a mole of B, so the rate of consumption of AB, -d[AB}/dt, is equal to the rate of formation of A, d[A]/dt, and the rate of formation of B, d[B]/dt.

However, what if you have AB2 --> A + 2B? You can see that every ONE mole of AB2 used produces TWO moles of B - in other words, B appears TWICE as fast as AB2 is used up. In this case, -d[AB2]/dt = d[A]/dt = (0.5) d[B]/dt. Notice how here you need to divide the rate of formation of B by two, since a reaction rate is expressed per ONE mole of substance.

Hope that clears things up for you! Let me know if you need any clarifications.