An experimental drug, D, is known to decompose in the blood stream. Tripling the concentration of the drug increases the decompo

Question

1 month ago

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sition 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.

1 answer:

lions [2.9K] · Answer 1 · 2026-03-05T07:52:45+03:00

Answer:

The rate law for the decomposition reaction is:

$R=k[D]^2$

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

Explanation:

$D\rightarrow Product$

The rate law can be expressed as:

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

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

$[D]'=3[D]$

$R'=9\times R$

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

[1] ÷ [2]

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

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

$9=3^x$

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

$R=k[D]^2$

The unit for the rate constant will be:

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

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