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2A → B + C
Two trials of the above reaction are run. The concentration of A in the second trial is three times that in the first. It is found that the initial rate of the reaction in the second trial is nine times the initial rate in the first trial. This indicates that the reaction is
1.zero order in [A]
2.first order in [A]
3.second order in [A]
4.third order in [A]

Answer :

Answer:

The correct answer is option 3.

Explanation:

2A → B + C

First trial ;

[A] = x , Rate of the reaction = R

Rate law of the reaction can be written as:

[tex]R=k[x]^a[/tex]..[1]

Second trial ;

[A] = 3x , Rate of the reaction = R' = 9R

Rate law of the reaction can be written as:

[tex]R'=k[3x]^a[/tex]

[tex]9R=k[3x]^a[/tex]..[2]

[1] ÷ [2]

[tex]\frac{R}{9R}=\frac{k[x]^a}{k[3x]^a}[/tex]

a = 2

Rate  of the reaction : [tex]R=k[A]^3[/tex]

The order of the reaction is 2. This is because in the rate law expression the the power of the concentration of reactant A id 2.

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