# Solution - Integrated Rate Equations - Half-life of a Reaction

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#### Question

(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

#### Solution

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Derive the relation between half life and rate constant for a first order reaction

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The experimental data for decomposition of N2O5

[2N_2O_5 -> 4NO_2 + O_2] in gas phase at 318K are given below:

 t(s 0 400 800 1200 1600 2000 2400 2800 3200 10^2xx[N_2O_5]mol L^(-1) 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35

(i) Plot [N2O5] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log [N2O5] and t.

(iv) What is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from and compare it with (ii).

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Calculate the half-life of a first order reaction from their rate constants given below:

(i) 200 s−1 (ii) 2 min−1 (iii) 4 years−1

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The rate constant for a first order reaction is 100 s–1. The time required for completion of 50% of reaction is _______.

(A) 0.0693 milliseconds

(B) 0.693 milliseconds

(C) 6.93 milliseconds

(D) 69.3 milliseconds

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The integrated rate equation for first order reaction is A → products

(a) 

(b) k=-1/tl_n[A]_t/[A]_0

(c) k=2.303/t log_10 ([A]_t/[A]_0 )

(d) k=l/tl_n[A]_t/[A]_0

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#### Reference Material

Solution for concept: Integrated Rate Equations - Half-life of a Reaction. For the courses 12th CBSE (Arts), 12th CBSE (Commerce), 12th CBSE (Science), 12th HSC Science (Computer Science), 12th HSC Science (Electronics), 12th HSC Science (General) , PUC Karnataka Science
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