Chemistry, asked by Dishabhtiwari, 9 months ago

Prove that the half life period of the zero order reaction is proportional to the

initial concentration of reactants?​

Answers

Answered by jdhanalaxmi1977
0

Answer

The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration (i.e. the time taken for the reactant concentration to reach half of its initial value). It is denoted by the symbol ‘t1/2’ and is usually expressed in seconds.

Half-Life Formula

It is important to note that the formula for the half-life of a reaction varies with the order of the reaction.

For a zero-order reaction, the mathematical expression that can be employed to determine the half-life is: t1/2 = [R]0/2k

For a first-order reaction, the half-life is given by: t1/2 = 0.693/k

For a second-order reaction, the formula for the half-life of the reaction is: 1/k[R]0

Where,

t1/2 is the half-life of the reaction (unit: seconds)

[R0] is the initial reactant concentration (unit: mol.L-1 or M)

k is the rate constant of the reaction (unit: M(1-n)s-1 where ‘n’ is the reaction order)

Derivation of Half-Life Formula for Zero-Order Reactions

For a zero-order reaction, the units of the rate constant are mol.L-1.s-1. The expression for a zero-order rate constant is:

k=[R]0−[R]t

Substituting t = t1/2, at which point [R] = [R]0/2 (at the half-life of a reaction, reactant concentration is half of the initial concentration).

k=[R]0−[R]0/2t1/2

Rearranging the equation, the expression for the half-life of a zero-order reaction is found to be:

t1/2=[R]02k

Derivation of Half-Life Formula for First-Order Reactions

For a first-order reaction, the rate constant can be mathematically expressed as follows:

k=2.303tlog[R]0[R]

From the definition of reaction half-life, at t = t1/2, [R] = [R]0/2. Substituting these values in the expression for the first-order rate constant, the following equation is obtained:

k=2.303t1/2log[R0][R]0/2

Rearranging the expression to find the value of t1/2:

t1/2=2.303klog(2)=0.693k

Thus, the half-life of a first-order reaction is given by 0.693/k.

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