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Half life of a chemical reaction at a particular concentration is 75 min. When the concentration of the reactant is doubled, the half life becomes 150 min. What is the order of reaction?
- a)First order
- b)Zero order
- c)Second order
- d)Third order
Correct answer is option 'B'. Can you explain this answer?
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Half life of a chemical reaction at a particular concentration is 75 ...
Given, half-life of a chemical reaction at a particular concentration = 75 min.
When the concentration of the reactant is doubled, the half-life becomes 150 min.
Let's assume the initial concentration of the reactant to be [A]₀ and its half-life to be t₁/₂.
We know that for a first-order reaction, the half-life is independent of the initial concentration of the reactant.
But, for a zero-order reaction, the half-life is directly proportional to the initial concentration of the reactant.
Let's find out the initial rate constant (k) for the given reaction using the formula:
t₁/₂ = 0.693/k
At initial concentration [A]₀,
t₁/₂ = 75 min, so k₁ = 0.693/75
Now, when the concentration of the reactant is doubled, the new concentration is 2[A]₀.
And, the new half-life (t₂/₂) = 150 min.
Using the same formula,
t₂/₂ = 0.693/k₂
Substituting the values,
150 = 0.693/k₂
k₂ = 0.693/150
We know that for a zero-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant.
So, if the initial concentration is doubled, the rate of the reaction will also get doubled.
Hence, the order of the reaction is zero-order.
Therefore, option B is the correct answer.
When the concentration of the reactant is doubled, the half-life becomes 150 min.
Let's assume the initial concentration of the reactant to be [A]₀ and its half-life to be t₁/₂.
We know that for a first-order reaction, the half-life is independent of the initial concentration of the reactant.
But, for a zero-order reaction, the half-life is directly proportional to the initial concentration of the reactant.
Let's find out the initial rate constant (k) for the given reaction using the formula:
t₁/₂ = 0.693/k
At initial concentration [A]₀,
t₁/₂ = 75 min, so k₁ = 0.693/75
Now, when the concentration of the reactant is doubled, the new concentration is 2[A]₀.
And, the new half-life (t₂/₂) = 150 min.
Using the same formula,
t₂/₂ = 0.693/k₂
Substituting the values,
150 = 0.693/k₂
k₂ = 0.693/150
We know that for a zero-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant.
So, if the initial concentration is doubled, the rate of the reaction will also get doubled.
Hence, the order of the reaction is zero-order.
Therefore, option B is the correct answer.
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Community Answer
Half life of a chemical reaction at a particular concentration is 75 ...
⇒ 2-1 = 2n-1
-1 = n - 1
⇒ n = 0
So it is zero order reaction.
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Half life of a chemical reaction at a particular concentration is 75 min. When the concentration of the reactant is doubled, the half life becomes 150 min. What is the order of reaction?a)First orderb)Zero orderc)Second orderd)Third orderCorrect answer is option 'B'. Can you explain this answer?
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