In the reaction: `ArarrB+C` , the rate of formation of C becomes twice if the concentration of A is made double. What is the order of the reaction ?

Explanation

If order of the reaction= n , then reaction -rate ,
`=-(d[A])/(dt)=(d[B])/(dt)=(d[C])/(dt)=k[A]^(n)`
`therefore (d[C])/(dt)=k[A]^(n)`
reaction -rate is doubled when concentration of A is doubled.
Hence, `2(d[C])/(dt)=k[2A]^(n)=2^(n)k[A]^(n)=2^(n)(d[C])/(dt)`
or, `2^(n)=2^(1) thereforen=1`