Question

An experimental drug, D, is known to decompose in the blood stream. Tripling the concentration of the drug increases the decomposition rate by a factor of nine. Write the rate law for decomposition of D and give the units of the rate constant in terms of seconds.

Answer 1

Answer:

The rate law of the decomposition reaction is :

$R=k[D]^2$

The unit of the rate constant  will be $M^{-1}s^{-1}$

Explanation:

$D\rightarrow Product$

The rate law can be written as';

$R=k[D]^x$..[1]

On tripling concentration of the drug increases the decomposition rate by a factor of nine.

$[D]'=3[D]$

$R'=9\times R$

$R'=k[D]'^x$...[2]

[1] ÷ [2]

$\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}$

$\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}$

$9=3^x$

On solving for x , we get;

x = 2

Second order reaction

The rate law of the decomposition reaction is :

$R=k[D]^2$

unit rate constant will be :

$k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}$

The unit of the rate constant  will be $M^{-1}s^{-1}$