#
SECOND ORDER
REACTION - HALF LIFE - SPECIFIC RATE - UNITS

# 1) For an irreversible reaction A ----> 2B, the rate is
increased by four times when the concentration of A is doubled. The
incorrect statement about this reaction is:

1) It is a second order reaction.

2) Half life is independent of initial concentration of A.

3) The unit of the specific rate, k is L mol^{-1 }s^{-1}.

4) Half life is inversely related to the
initial concentration of A.

**Logic:**

Except one, the remaining three statements are correct regarding
the given reaction. The logic is to find out the order of reaction
comparing the rates at different concentrations.

## **Solution:**

The rate law for a given reaction reaction can be written as
follows:

where:

k = specific rate

[A] = concentration of reactant, A

n = order of reaction

Let

the initial rate: r_{1} = k[A_{1}]^{n}

the rate after doubling the conc. of A: r_{2}
= k[A_{2}]^{n}

Also we know that:

[A_{2}] = 2[A_{1}] (since the
concentration of A is doubled)

and

r_{2} = 4r_{1 } (since the rate is
increased by four times)

Now by comparing rates at two different conditions, it is possible to find
the order of the reaction.

or

4 = 2^{n}

Therefore:

n = 2

That means given reaction is a second order
reaction.

* Now we know that for a second order
reaction, the half life is inversely proportional to the initial
concentration of the reactants and the units of specific rate can be
written as: L mol^{-1}^{ }s^{-1}.

In general:

unit for rate constant or specific rate, k is mol^{1-n} L^{n-1
}time^{-1}.

Conclusion:

Since it is a second order reaction, options 1,3 and 4 are correct
statements.

### Homework:

1) What should be the order of reaction if the half life is independent of
initial concentration of reactants?

2) What are the units of zero, first and third order reactions?

3) What will be the order of reaction if the rate is doubled when
concentration of reactant is doubled?