The given reaction is A---> products
Trial-1: The initial concentration of A is 0.10 M when the initial rate is 0.015 M/s
Trial-2: The initial concentration of A is 0.40 M when the initial rate is 0.060 M/s
Let the order be m
Rate law for first trial will be: [tex]0.015M/s = k.[0.10M]^{m}[/tex]
Rate law for second trial will be: [tex]0.060M/s = k.[0.40M]^{m}[/tex]
Rate- 2/ Rate-1 : [tex]\frac{0.060M/s}{0.0150M/s}=\frac{k(0.40)^{m}}{k.(0.10)^{m}}[/tex]
[tex]4 = 4^{m}[/tex]
So, m = 1
Therefore the order of the reaction is 1
