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__ __∆[NH____3____]__ = __-1__ __∆[O____2____]__ = +__1 __ __∆ [NO]__ = +__1__ __∆[H____2____O]__

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