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Assoli18
3 months ago
13

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

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.
Chemistry
1 answer:
lions [2.9K]3 months ago
5 0

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}.

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