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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.
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Topics: Electrochemistry and Chemical Kinetics
(87)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Treatment of acetaldehyde with ethyl magnesium bromide and subsequent hydrolysi
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- Choose isosteres from the following.
Topics: Electrochemistry and Chemical Kinetics (87)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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