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Sliva [168]
3 years ago
8

In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.

Chemistry
1 answer:
mihalych1998 [28]3 years ago
8 0

In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol t_{1/2}.

(a) For first order reaction, rate constant and half life time are related to each other as follows:

k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}

Thus, rate constant of the reaction is 0.066 min^{-1}.

(b) Rate equation for first order reaction is as follows:

k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}

now, 75% of the compound is decomposed, if initial concentration [A_{0} ] is 100 then concentration at time t [A_{t} ] will be 100-75=25.

Putting the values,

0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)

On rearranging,

t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min

Thus, time required for 75% decomposition is 21 min.

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