Show that the time required for the completion of 75% of a reaction of first order is twice the time required for the completion of 50% of the reaction.

For space the space first space order space reaction straight t equals space fraction numerator 2.303 over denominator straight K end fraction log straight A subscript 0 over straight A straight t subscript 1 space equals fraction numerator 2.303 over denominator straight K end fraction log fraction numerator space 100 over denominator left parenthesis 100 minus 75 right parenthesis end fraction space equals fraction numerator space 2.303 over denominator straight K end fraction log space 100 over 25 space.....1 straight t subscript 2 space equals space fraction numerator 2.303 over denominator straight K space end fraction log fraction numerator 100 over denominator left parenthesis 100 minus 50 right parenthesis end fraction equals fraction numerator 2.303 over denominator straight K end fraction log 100 over 50 space.........2 Eq. space left parenthesis 1 right parenthesis space divided space by space eq space 2 straight t subscript 1 over straight t subscript 2 space equals 4 over 2 comma space straight t subscript 1 space equals space 2 straight t subscript 2

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Chemical Kinetics

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