For the reaction Zero order
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Explanation
$ t _ { 1/2} \alpha C_o $
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For reaction first order
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Explanation
$ t_ { 1/2} = { 0.693 \over k } $
Which of the following represents the expression for ¾th life of a first order reaction
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Explanation
$ {2.303 \over k } log 4 $ $ t_{3/4} = { 2.303 \over k} log { ao \over ao -ao \times {3 \over 4} } = { 2.303 \over k } log { ao \over { ao /4 } } $
If initial concentration is doubled, the time for half reaction is also doubled. The order of reaction is ...
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Explanation
Zero for zero order reaction $ t_ { 1/2} \alpha C_o $
If a is the initial concentration of the reactant, the half life period of the reaction of the $n^{th} $ order is proportional to ...
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Explanation
$ a^ {1-n} t_{1/2} \alpha a^ {1-n} $
For the first order reaction, half life is 14 s. The time required for the initial concentration to reduce to 1/8th of its value is ...
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Explanation
42 S $ Ao \rightarrow {Ao \over 2} \rightarrow { Ao \over 4} \rightarrow { Ao \over 8} $ $ \therefore 3 \times t _ {1/2} $
In the first order reaction the concentration of the reactants is reduced to 25% in one hour. The half life period of the reaction is …
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Explanation
$ 30 min 100 \% \rightarrow 50 \rightarrow 25 \% $ $ \therefore T = 2 \times t {1/2} \therefore t{1/2} = { T \over 2 } $
For the First order reaction with half life is 150 seconds, the time taken for the concentration of the reactant to fall from m/10 to m/100 will be approximately
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Explanation
$ 500S \rightarrow { M \over 10 } \rightarrow { M \over 20 } \rightarrow {M \over 40 } \rightarrow { M \over 80} \rightarrow { M \over 180 } $ $ \therefore T \cong 3 \times t {1/2} to 4 \times t {1/2} \cong 450 to 600 S $
For the reaction $N_2O_5 --> 2NO_2 + ½O_2 t_{½ }= 24 hrs $ . starting with 10 g of $N_2O_5$ how many grams of $N_2O_5$ will remain after a period of 96 hours ?
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Explanation
$ 0.63 g T = n \times {1/2} n = { 96 \over 24 } = 4 $ $ \therefore Amount left = {a \over 2 ^ n } = { 10 \over 2^ 4 } = { 10 \over 16 } = 0.63 $
In the first order reaction 75% of reactant disappeared in 1.386 h. Calculate the rate constant of reaction.
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Explanation
$ 2.8 \times 10 ^ {-4} S^ {-1} K = { 2.303 \over 1.386 \times 60 \times 60 } log { a \over a -0.75 a } = 2.8 \times 10^ {-4} S^{-1} $
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