Question

When initial concentration of the reactant is doubled , the half-life period of a zero order reaction

• A. remains unchanged
• B. is halved
• C. is tripled
• D. is doubled

Answer 1

The Correct Option is D

Half life of zero order

$t _{1 / 2}=\frac{\left[ A _{0}\right]}{2 K }$
$t _{1 / 2}$ will be doubled on doubling the initial concentration.

Answer 2

First of all, let us see what the order of a reaction is.

The order of a reaction is basically the sum of the powers of the reactant concentration in the rate law expression. The order of a reaction can be 0, 1, 2, 3, and also a fraction.

For a 0-order reaction, the rate of reaction is proportional to the 0 power of the reactant concentration.

R→ P

Rate = \(\frac{-d[R]}{dt}\)K[R]⁰

A 0-order reaction is the decomposition of gaseous ammonia on hot platinum.

2NH₃(g)  1130K Pt→  N₂(g)+3H₂(g)

⇒ Rate = K[NH₃]0 = k

Now let us talk about what half-life is (t_1/2).

When the concentration of the reactant gets lower than half its original concentration, this is known as the half-life.

In a 0-order reaction, the half-life is:

t_1/2 = \(\frac{[R0]}{2k}\) ……..(1)

Where [R₀] is the initial concentration of the reactant.

According to the question, the initial concentration of the reactant is 2x and we are looking at the new half-life of the 0-order reaction.

Initially, when the concentration of the reactant is [R₀], the half-life will be:

t_1/2 = \(\frac{[R0]}{2k}\)………(2)

After the concentration of the reactant is doubled:

[R₀]’ = 2[R₀] ..........(3)

Now substitute equation (3) in equation (1)

t_1/2’ = \(\frac{[R0]}{2k}\)

⇒ t_1/2’ = 2×\(\frac{[R0]}{2k}\)

From (2): \(\frac{t1}{2}\)^’ = 2×\(\frac{t1}{2}\) 

Hence we can say that after doubling the concentration of the reactant, the half-life period of this zero-order reaction will be doubled.

 

So, the correct answer is “Option D”.