What is the order of reaction for which rate becomes half if volume of the container having same amount of reactant is doubled? Assume gaseous phase reaction.

Answer

1

Explanation

Rate,
For `I`: Let a mole of reactant in vessel of `V` litre
`:. r_(1)=K[a/V]^(n) …(1)`
For `II`: The volume is doubled, rate becomes half.
`:. R_(1)/2=K[a/(2V)]^(n) …(2)`
By eqs. (1) and (2)
`2=2^(n)` or `n=1`
Order of reaction =`1`