The rate of a reaction can be expresses in terms of a change in the amount of reactants or products over an interval of time.
Let's take a look at the reaction A --> 2B
Before we write the rate expression, let's go over some RULES.
Rate of change of reactants is always negative.
Rate of change of products is always positive.
In the rate expression, we must do 1/coefficient for each reactant and product.
We express rates in terms of ∆concentration/∆t
A --> 2B
Rate = -∆[A] = +1 ∆[B]
∆t 2 ∆t
The coefficient in from of A is 1, so we would do 1/1 which we don't have to write. A is a reactant so we have to put a minus sign since as it reacts, its concentration would decrease. We also write∆[A]/∆t which means the change in concentration of A over change in time. B is a product so it would be increasing, positive sign. The coefficient in front of B is 2, so we have to write 1/2.
Some terminology: ∆[A] /∆t is the rate of disappearance of A and ∆[B] /∆t is the rate of appearance of B.
Example: Write the expression for the following reaction in terms of the reactants and of the products:
4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O (g)
Rate = -1 ∆[NH3] = -1 ∆[O2] = +1 ∆ [NO] = +1 ∆[H2O]
4 ∆t 5 ∆t 4 ∆t 6 ∆t
Example: Given the rate of disappearance of D = 2.5*10^-2 M/s
Calculate the rate of disappearance of E according to the following equation:
D+3E --> 2F
Let's begin by writing rate expression in terms of D and E. They are both reactants so should both have a negative sign.
-∆[D] = -1 ∆[E]
∆t 3 ∆t
Now we can plug in 2.5*10^-2 M/s for ∆[D]/∆t (the rate of disappearance of D )
We get - 2.5*10^-2 M/s = -1 ∆[E]
3 ∆t
∆[E] = 7.5*10^-2 M/s
∆t