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A graph plotted between log ⁡ t 50 vs log concentration is a straight line. What conclusion can you draw from the given graph? [Given: n = order of reaction]

Options

  1. An   =   1,   t 1 2 = 1 k .a
  2. Bn = 2 , t 1 2 = 1 a
  3. Cn   =   1,   t 1 2   =   0 .693 k
  4. DNone of the above

Correct answer

C. n   =   1,   t 1 2   =   0 .693 k

Step-by-step solution

t 1 2 ∝ a 1 - n or t 1 2 = Z a 1 - n or log ⁡ t 1 2 = log ⁡ Z + ( 1 - n ) log ⁡ a or ( y ) = c + m x Thus, slope = m = 1 - n or 1 - n = 0 So n = 1 and for I order reaction t 1 2 = 0.693 k .

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