If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is

Options

(a) zero
(b) fraction
(c) three
(d) one

Correct Answer:

zero

Explanation:

We know for a reaction of n th order. k = 1/(n-1)t [1/(a-x)ⁿ⁻¹ – 1/aⁿ⁻¹].
when t=t(1/2), a-x = a/2.
t(1/2) = 1/(n-1)K [ 1/(a/2)ⁿ⁻¹ – 1/aⁿ⁻¹] …(i).
This time for t(1/2).
When a=2a, then time period for half reaction(t) is given by,
t = 1/(n-1)k [1/(a)ⁿ⁻¹ -1/(2a)ⁿ⁻¹]..(ii).
But t = 2t(1/2) , from (i) and (ii)
t(1/2)/t = [1/(a/2)ⁿ⁻¹ – 1/aⁿ⁻¹] / (1/a)ⁿ⁻¹ – 1/(2a)ⁿ⁻¹ .
1/2 = aⁿ⁻¹ – (a/2)ⁿ⁻¹/(a/2)ⁿ⁻¹ ⨯ (2a)ⁿ⁻¹ / (2aⁿ⁻¹- aⁿ⁻¹).
1/2 = (a/2)ⁿ⁻¹ (2ⁿ⁻¹ -1) / (a/2)ⁿ⁻¹ ⨯ (2a)ⁿ⁻¹/aⁿ⁻¹ (2ⁿ⁻¹ – 1).
1/2 = (2)ⁿ⁻¹ ⇒ n = 0. Thus order of the reaction is zero.