For a first order reaction A → B , the rate constant, k = 5 . 5 × 10 - 14 s - 1 . The time required for 67 % completion of reaction is x × 10 - 1 times the half life of reaction. The value of x is Nearest integer) (Given : l

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For a first order reaction A &#8594; B , the rate constant, k = 5 . 5 &#215; 10 - 14 s - 1 . The time required for 67 % completion of reaction is x &#215; 10 - 1 times the half life of reaction. The value of x is Nearest integer) (Given : log 3 = 0 . 4771 )

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For first order reaction, t = 2 . 303 k log A 0 A t Now, t 67 % = 2 . 303 k log A 0 . 33 A & t 50 = 2 . 303 k log A 0 . 5 A t 67 % = 1 k ln 1 1 - 0 . 67 = t 1 / 2 ln 2 &#215; ln 1 1 - 2 3 &#8658; t 67 % = 1 . 585 &#215; t 1 / 2 X &#215; 10 - 1 = 1 . 585 &#8658; X = 15 . 85

For a first order reaction A → B , the rate constant, k = 5 . 5 × 10 - 14 s - 1 . The time required for 67 % completion of reaction is x × 10 - 1 times the half life of reaction. The value of x is Nearest integer) (Given : log 3 = 0 . 4771 )

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For a first order reaction A &#8594; B , the rate constant, k = 5 . 5 &#215; 10 - 14 s - 1 . The time required for 67 % completion of reaction is x &#215; 10 - 1 times the half life of reaction. The value of x is Nearest integer) (Given : log 3 = 0 . 4771 )

Correct answer

Step-by-step solution

More from Chemical Kinetics

For a first order reaction A → B , the rate constant, k = 5 . 5 × 10 - 14 s - 1 . The time required for 67 % completion of reaction is x × 10 - 1 times the half life of reaction. The value of x is Nearest integer) (Given : log 3 = 0 . 4771 )