Answer:
The rate law of the decomposition reaction is :
[tex]R=k[D]^2[/tex]
The unit of the rate constant will be [tex]M^{-1}s^{-1}[/tex]
Explanation:
[tex]D\rightarrow Product[/tex]
The rate law can be written as';
[tex]R=k[D]^x[/tex]..[1]
On tripling concentration of the drug increases the decomposition rate by a factor of nine.
[tex][D]'=3[D][/tex]
[tex]R'=9\times R[/tex]
[tex]R'=k[D]'^x[/tex]...[2]
[1] ÷ [2]
[tex]\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}[/tex]
[tex]\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}[/tex]
[tex]9=3^x[/tex]
On solving for x , we get;
x = 2
Second order reaction
The rate law of the decomposition reaction is :
[tex]R=k[D]^2[/tex]
unit rate constant will be :
[tex]k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}[/tex]
The unit of the rate constant will be [tex]M^{-1}s^{-1}[/tex]
The units of the rate constant in terms of seconds for a second order reaction is M-1s-1.
The rate of reaction refers to how quickly or slowly a reaction proceeds. The rate law is obtained from the concentration of the reactants Hence, the rate rate law is obtained as; Rate = k[A]^x
Where;
k = rate constant
[A] = concentration of reactants
x = order of reaction
If tripling the concentration of the drug increases the decomposition rate by a factor of nine then the reaction is second order. The units of the rate constant in terms of seconds for a second order reaction is M-1s-1.
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