Negative Rate Constant?
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Negative Rate Constant?
Is it possible to have a negative rate constant k? If so, how would that affect the rate of the reaction?
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Re: Negative Rate Constant?
Rate constant k should always be positive.
From the Arrhenius Equation, we know k = A x exp(-Ea/RT).
"A" (frequency factor) will always be positive because (according to Google) there are no experimental cases where A is negative, and mathematically exp(-Ea/RT) can never be negative.
However, if the reaction is 2nd order, we know that the slope of the straight line plot of 1/[A] vs time is -k.
From the Arrhenius Equation, we know k = A x exp(-Ea/RT).
"A" (frequency factor) will always be positive because (according to Google) there are no experimental cases where A is negative, and mathematically exp(-Ea/RT) can never be negative.
However, if the reaction is 2nd order, we know that the slope of the straight line plot of 1/[A] vs time is -k.
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Re: Negative Rate Constant?
Another way to think of it is that reactions rates are always positive and since k is a proportionality constant that relates some given concentration(s) (which are always positive) with the rate (also always positive), it's not possible for k to be negative.
Last edited by Neil DSilva 1L on Tue Feb 24, 2015 2:57 am, edited 2 times in total.
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Re: Negative Rate Constant?
Tatyana Diaz wrote:Rate constant k should always be positive.
From the Arrhenius Equation, we know k = A x exp(-Ea/RT).
"A" (frequency factor) will always be positive because (according to Google) there are no experimental cases where A is negative, and mathematically exp(-Ea/RT) can never be negative.
However, if the reaction is 2nd order, we know that the slope of the straight line plot of 1/[A] vs time is -k.
I'd like to correct the last statement of my post!
If rxn is 2nd order, line plot has a slope of +k.
If rxn is 1st or 0 order, line plot has a slope of -k.
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Re: Negative Rate Constant?
Also, in first/zero order reactions, as Tatyana mentioned, the straight line plot has slope -k, but we have to add the negative sign because k has to remain positive, but the line has a downward slope.
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