1.

The half-life of a reaction is halved as the initial concentration of the reactant is doubled. The order of reaction is

Answer»

`0.5`
1
2
0

Solution :`t_(1//2) PROP (1)/([A_(0)]^(n-1)), ((t_(1//2))1)/((t_(1//2))2) = ([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1)) = {([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)`
`(t)/(t//2) = (2a)/(a)^(n-1) or 2 = (2)^(n-1)`
or ` n -1 = 1 or n =2 `


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