Following is the graph between `log T_(50)` and `log a` (`a =` initial concentration) for a given reaction at `27^(@)C`. Hence order is A. 1B. 2C. 3D. 0

Following is the graph between `log T_(50)` and `log a` (`a =` initial

1.	Following is the graph between `log T_(50)` and `log a` (`a =` initial concentration) for a given reaction at `27^(@)C`. Hence order is A. 1B. 2C. 3D. 0
Answer» Correct Answer - D `t_(1//2)prop((1)/(a))^(n-1)" or "t_(1//2)=k(a)^(1-n)` `logt_(1//2)=logk+(1-n)loga` (It represents straight line equation, y = c + mx) Slope `=(1-n)=tan45^(@)=1` `(1-n)=1implies n=0`

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Following is the graph between `log T_(50)` and `log a` (`a =` initial concentration) for a given reaction at `27^(@)C`. Hence order is A. 1B. 2C. 3D. 0

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