Answer :
Answer: The rate of the reaction gets doubled when the concentration of the reactant is doubled.
Explanation:
We are given:
Order of the reaction = 1
Rate law equation for any order follows:
[tex]\text{Rate}=k[A]^n[/tex]
where, n = order of the reaction
[tex][A][/tex] = concentration of the reactant A
Rate law equation for the first order reaction becomes:
[tex]\text{Rate}_1=k[A][/tex] .........(1)
When the concentration of the reactant is doubled, the new rate law becomes:
[tex]\text{Rate}_2=k[2A]\\\\\text{Rate}_2=2k[A][/tex] .....(2)
Taking the ratio of equation 2 and equation 1, we get:
[tex]\frac{\text{Rate}_1}{\text{Rate}_2}=\frac{2k[A]}{k[A]}\\\\\text{Rate}_2=2\times \text{Rate}_1[/tex]
Hence, the rate of the reaction gets doubled when the concentration of the reactant is doubled.
The rate will gets doubled, when the concentration is doubled.
What is first order reaction?
A first-order reaction depends on the concentration of one reactant, and the rate law is:
Rate = k[A]
Given:
The concentration is doubled.
As we know, rate is dependent on the concentration in first order reaction. Thus on doubling the concentration, the rate will gets doubled too.
Find more information about First order reaction here:
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