A reaction is of second order with respect to a reactant . How is its effected if the concentration of the reactant is (i)doubled (ii) reduced to half ?

Options

A. Becomes 2 times and `(1)/(2)` times respectively
B. Becomes` (1)/(2) "timses and" (1)/(2)` times respectively
C. Becomes 2 times and 4 times, respectively
D. Becomes 4 time and 2 times, respectively

Explanation

(For first order reaction ), rate =k [A] we have to find the rate expression when conecentration is doubled or reduced to half and compare it with the normal rate. E.g.
for a reaction A `to` products
Rate =k[A]=ka
(i) When the concentration of A is doubled
[A]=2a
Rate =k (2a)=k2a
Rate of reaction becomes 2 times (ii) When concentration of Ais reduced to `(1)/(2)`
` [A] =(1)/(2)a`
Rate = `k ((a)/(2)) =(ka)/(2)`
Rate of reaction becomes `(1)/(2)` times i.e reduced to twice