**First order reaction**

– Let us consider a first order reaction:

**A → products**

– Suppose that at the beginning of the reaction (*t* = 0), the concentration of A is a moles liter^{–1}.

– If after time t, x moles of A have changed, the concentration of A is *a* – *x*.

– We know that for a first-order reaction, the rate of reaction, d*x*/d*t*, is directly proportional to the concentration of the reactant. Thus,

– Integration of the expression (1) gives:

– where (I) is the constant of integration.

– The constant *k* may be evaluated by putting *t* = 0 and *x* = 0. Thus,

I = – ln *a*

– Substituting for (I) in equation (2):

– Changing into common logarithms:

– The value of *k* can be found by substituting the values of a and (*a * – *x*) determined experimentally at time interval *t* during the course of the reaction.

– Sometimes the integrated rate law in the following form is also used:

where* x*_{1} and *x*_{2} are the amounts decomposed at time intervals *t*_{1} and *t*_{2} respectively from the start.

**Examples of First order reactions**

Some common reactions which follow first order kinetics are listed below :

**(1) Decomposition of N _{2}O_{5} in CCl_{4} solution**

– Nitrogen pentoxide in carbon tetrachloride solution decomposes to form oxygen gas,

**N _{2}O_{5} → 2NO_{2} + 1/2 O_{2}**

– The reaction is carried out in an apparatus shown in the following Figure:

– The progress of the reaction is monitored by measuring the volume of oxygen evolved from time to time.

– If *V*_{t} be the volume of O_{2} at any time t and *V*_{∞} the final volume of oxygen when the reaction is completed, the *V*_{∞} is a measure of the initial concentration of N_{2}O_{5} and (V_{∞} – V_{t}) is a measure of undecomposed N_{2}O_{5} (*a* – *x*) remaining at time *t*. Thus,

– On substituting values of *V*_{∞ }, (*V*_{∞} – *V*_{t}) at different time intervals, *t*, the value of* k* is found to be constant. Thus it is a reaction of the first order.

**Solved Problem (1):** From the following data for the decomposition of N_{2}O_{5} in CCl_{4} solution at 48°C, show that the reaction is of the first order

Solution:

– For a first order reaction, the integrated rate equation is:

– In this example, V_{∞} = 34.75

Since the value of k is fairly constant, it is a first order reaction.

**(2) Decomposition of H**_{2}O_{2} in aqueous solution

_{2}O

_{2}in aqueous solution

– The decomposition of H_{2}O_{2} in the presence of Pt as catalyst is a first order reaction.

– The progress of the reaction is followed by titrating equal volumes of the reaction mixture against standard KMnO_{4} solution at different time intervals.

**Solved Problem (2):** A solution of H_{2}O_{2} when titrated against KMnO_{4} solution at different time intervals gave the following results :

Show that the decomposition of H_{2}O_{2} is a first order reaction.

Solution:

– The integrated rate equation for first order reaction is

Since the volume of KMnO_{4} used in the titration is the measure of the concentration of H_{2}O_{2} in the solution

– Substituting these values in the rate equation above, we have;

– Since the value of k is almost constant, the decomposition of H_{2}O_{2} is a first order reaction.

**(3) Hydrolysis of an Ester**

– The hydrolysis of ethyl acetate or methyl acetate in the presence of a mineral acid as catalyst, is a first order reaction.

**CH _{3}COOC_{2}H_{5} + H_{2}O → CH_{3}COOH + C_{2}H_{5}OH**

– For studying the kinetics of the reaction, a known volume of ethyl acetate is mixed with a relatively large quantity of acid solution, say N/2 HCl.

– At various intervals of time, a known volume of the reaction mixture is titrated against a standard alkali solution.

– Hydrolysis of the ester produces acetic acid.

– Therefore as the reaction proceeds, the volume of alkali required for titration goes on increasing.

**Solved Problem (3):** The following data was obtained on hydrolysis of methyl acetate at 25°C in 0.35N hydrochloric acid. Establish that it is a first order reaction.

Solution:

– For a first order reaction:

– At any time, the volume of alkali used is needed for the acid present as catalyst and the acid produced by hydrolysis.

– The volume of alkali used for total change from t_{0} to t_{∝} gives the initial concentration of ester. Thus,

– Substituting values in the rate equation above, we have:

Since the values of k in the two experiments are fairly constant, the reaction is of the first order.

**(4) Inversion of Cane sugar (sucrose). **

– The inversion of cane sugar or sucrose catalyzed with dil HCl to give D-glucose and D-fructose follows the first order kinetics

**C _{12}H_{22}O_{11} +H_{2}O ⎯⎯→ C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6}**

– The progress of the reaction is followed by noting the optical rotation of the reaction mixture with the help of a polarimeter at different time intervals.

– The optical rotation goes on changing since D-glucose rotates the plane of polarised light to the right and D-fructose to the left.

– The change in rotation is proportional to the amount of sugar decomposed.

– Let the final rotation be *r*_{∞}, the initial rotation *t*_{0} while the rotation at any time *t* is *r*_{t}

The initial concentration, a is ∞ ( *t*_{0} – *r*_{∝}).

The concentration at time t, (*a* – *x*) is ∝ (*r*_{t} – *r*_{∝})

– Substituting in the first order rate equation,

– If the experimental values of *t* ( *t*_{0} – *r*_{∝}) and (*r*_{t} – *r*_{∝}) are substituted in the above equation, a constant value of *k* is obtained.

**Solved Problem (4):** The optical rotation of sucrose in 0.9N HCl at various time intervals is given in the table below.

Show that inversion of sucrose is a first order reaction.

Solution:

– The available data is substituted in the first order rate equation for different time intervals

*r _{0}* –

*r*

_{∝}= 24.09 – (– 10.74) = 34.83 for all time intervals. Thus, the value of rate constant can be found.

Since the value of k comes out to be constant, the inversion of sucrose is a first order reaction.

**Units of First order Rate constant**

The rate constant of a first order reaction is given by:

– Thus the rate constant for the first order reaction is independent of the concentration.

– It has the unit **time ^{–1}**

**Calculation of Half-life of a First order Reaction**

– Reaction rates can also be expressed in terms of half-life or half-life period.

– It is defined as **the ****time required for the concentration of a reactant to decrease to half its initial value**.

– In other words, half-life is the time required for one-half of the reaction to be completed.

– It is represented by the symbol t_{1/2} or t_{0.5}.

– The integrated rate equation (4) for a first order reaction can be stated as :

where [A]_{0} is initial concentration and [A] is concentration at any time t. Half-life, t_{1/2}, is time when initial concentration reduces to 1/2 i.e.,

Substituting values in the integrated rate equation, we have:

It is clear from this relation that :

(1) The half-life for a first order reaction is independent of the initial concentration.

(2) it is inversely proportional to k, the rate-constant.