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Question

Explain half-life for zero-order and second-order reactions


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Solution

The half-life of Zero-order:

  • The zero-order reaction refers to a condition in which the rate of the reaction does not vary with the change in concentration of the reactant.
  • Half-life refers to the amount of time required to reduce the amount of a substance to half of its initial value.

Let. AB+C;

The rate law is given as:

Rate = k[A]n; where n is the order of the reaction. for zero-order n=0

The differential rate law is given as: k=-d[A]0dt

rearranging: d[A]0=-kdt

Integrating both side: [A]0[A]d[A]=-0tkdtA[A]0[A]=-kt[A]-[A]0=-kt[A]=[A]0-kt----->(i)

For half-life: at t=t12; [A]=[A]02

Substituting in eq(i)

[A]02=[A]0-kt12t12=[A]02k.

The half-life for second-order:

Let, 2AB

the differential rate law is given as: -d[A]dt=k[A]2

rearranging and integrating.

d[A][A]2=-kdt[A]0[A]d[A][A]2=-0tkdt[-1[A]][A]0[A]=-kt1[A]-1[A]0=kt1[A]=kt+1[A]0

For half-life: at t=t12; [A]=[A]02

substituting in the above equation

1[A]02=kt12+1[A]0t12=1k[A]0


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