Following is the graph between log `t_(1//2)` and log a (a initial concentration ) for a given reaction at `27^(@)C` Find the order of reaction

Following is the graph between log `t_(1//2)` and log a (a initial con

1.	Following is the graph between log `t_(1//2)` and log a (a initial concentration ) for a given reaction at `27^(@)C` Find the order of reaction
Answer» `t_(1//2) prop (1)/(a_(0)^(n-1)) implies t_(1//2) k = (1)/(a_(0)^(n-1))` log `t_(1//2)` = log k - (n-1) log `a_(0)` tan `45^(@) = -(n-1) = -(0-1)` tan `45^(@) = 1 " " therefore n = 0`

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Following is the graph between log `t_(1//2)` and log a (a initial concentration ) for a given reaction at `27^(@)C` Find the order of reaction

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