The half-life of the first-order decomposition of N H 2 N O 2 is 6.93 h at 1 2 ∘ C . If 6.2 g is allowed to decompose, how long does 99% decomposition take in hours?

The fraction left after 99% decomposition is 0.01 . Thus t = ln 2 t 1/2 ln ( 100 ) = ln 2 6.93 ln 100 = 46.04 h , Using the stated 6.93-hour half-life and full-precision logarithms gives 46.0419 … h, or 46.04 h to two decimal places.

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ChemistryChemical KineticsFirst Order ReactionsMedium2 minQB

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The half-life of the first-order decomposition of $N H_{2} N O_{2}$ is $6.93 h$ at $1 2^{\circ} C$ . If $6.2 g$ is allowed to decompose, how long does $99%$ decomposition take in hours?

Answer:

46.04

Solution:

The fraction left after $99%$ decomposition is $0.01$ . Thus

$t = \frac{t _{1/2}}{ln 2} ln (100) = \frac{6.93 ln 100}{ln 2} = 46.04 h,$

Using the stated 6.93-hour half-life and full-precision logarithms gives $46.0419 \dots$ h, or $46.04$ h to two decimal places.

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:First Order Reactions

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