Concentration and rate in first order reactions? Watch

Kurraiyo
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In first order reactions, rate=k[reactant] so as the concentration of the reactant increases the rate increases as well.

In my chemistry book it says, "Radioactive decay is an example of a first order reaction because its rate is independent of the concentration of the radioactive material."

:confused: How is that so?
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Girling1996
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With radioactive decay, there is half-life which is always constant, i.e if the halflife is 2 years and you have 100g of it.
after 2 years, there will be 50g, after 4 years 25g, etc.
Radioactive decay is first order because the rate is always the same.
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Kurraiyo
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(Original post by Girling1996)
With radioactive decay, there is half-life which is always constant, i.e if the halflife is 2 years and you have 100g of it.
after 2 years, there will be 50g, after 4 years 25g, etc.
Radioactive decay is first order because the rate is always the same.
So if you start with 100g or 1000g of a radioactive material, the RATE of decay would remain the same? I know the half-life would but the book says rate too :|
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Goods
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(Original post by Kurraiyo)
In first order reactions, rate=k[reactant] so as the concentration of the reactant increases the rate increases as well.

In my chemistry book it says, "Radioactive decay is an example of a first order reaction because its rate is independent of the concentration of the radioactive material."

:confused: How is that so?
Your chemistry text book is wrong. The number of radioactive decays is directly proportional to the number of unstable nuclei (the amount of radioactive material) and its the fact that it is directly proportional to the amount and not the square of the amount etc... that makes it first order. In the same way a first order rate equation only depends on the concentration not its square etc..
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Girling1996
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(Original post by Kurraiyo)
So if you start with 100g or 1000g of a radioactive material, the RATE of decay would remain the same? I know the half-life would but the book says rate too :|
I believe so, rate is always constant if it is first order.
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Goods
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(Original post by Girling1996)
I believe so, rate is always constant if it is first order.
the rate varies, its the rate of change of the rate (second derivative) that remains constant in a first order reaction.

Rate=\frac{\mathrm{d} [Conc]}{\mathrm{d} t}=k[Conc]

\frac{\mathrm{d}[Rate] }{\mathrm{d} t}=\frac{\mathrm{d}^{2} [Conc]}{\mathrm{d} t^{2}}=k
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Kurraiyo
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(Original post by Goods)
Your chemistry text book is wrong. The number of radioactive decays is directly proportional to the number of unstable nuclei (the amount of radioactive material) and its the fact that it is directly proportional to the amount and not the square of the amount etc... that makes it first order. In the same way a first order rate equation only depends on the concentration not its square etc..
Okay that makes sense, so only the half-life remains constant. Thank you!
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Borek
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Note: there do exist zeroth order reactions, in which reaction rate doesn't depend on the concentration at all.

Radioactive decay is definitely not one of them.
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