NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
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
- An   =   1,   t 1 2 = 1 k .a
- Bn = 2 , t 1 2 = 1 a
- Cn   =   1,   t 1 2   =   0 .693 k
- 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 .