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In the first order reaction concentration of reactant decreased from 0.4M to 0.1M in 20 minutes then rate constant of reaction will be: 0.693 /min 6.93 /min 0.0693 /min 69.3 /min in the given solution, how did they arrive at  straight t subscript 3 divided by 4 end subscript ? And can we solve this question by some other method? SOLUTION: t subscript 1 divided by 2 end subscript equals 1 half cross times t subscript 3 divided by 4 end subscript space equals space 1 half cross times 20 space equals space 10 space m i n t subscript 1 divided by 2 space end subscript equals space fraction numerator 0.693 over denominator K end fraction space equals space 10 space m i n K equals 0.0693 space m i n to the power of negative 1 end exponent  
Asked by g_archanasharma | 20 Mar, 2019, 19:20: PM
answered-by-expert Expert Answer
H e r e space t subscript 3 divided by 4 end subscript space r e p r e s e n t s space t h e space t i m e space r e q u i r e d space t o space c h a n g e space c o n c e n t r a t i o n space f r o m space 0.4 M space t o space 0.1 M space w h i c h space i s space 3 over 4 t h space c o n v e r s i o n A l t e r n a t i v e l y space w e space c a n space u s e space r e l a t i o n space f o r space 1 s t space o r d e r space r e a c t i o n space w h i c h space i s space minus t space space equals space fraction numerator 2.303 over denominator k end fraction log fraction numerator a over denominator a minus x end fraction space w h e r e space a space i s space i n i t i a l space c o n c e n t r a t i o n space a n d space left parenthesis a minus x right parenthesis space i s space f i n a l space c o n c e n t r a t i o n.
Answered by Sumit Chakrapani | 22 Mar, 2019, 00:31: AM

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