## If slope for ln(A) vs. t is negative is k?

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Nicole Leppi 2K
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### If slope for ln(A) vs. t is negative is k?

For example if a slope for a first order reaction (ln(A)) vs. t) was -3 would k=-3 or would k=3?

Nicole Leppi 2K
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### Re: If slope for ln(A) vs. t is negative is k?

that's what I thought, thanks!

Justin Le 2I
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### Re: If slope for ln(A) vs. t is negative is k?

Since the slope is equal to -k and the slope is -3, then k = 3. k is always positive.

martha-1I
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### Re: If slope for ln(A) vs. t is negative is k?

Rate constants (k) are always positive.
Rates can be negative or positive depending on whether it's for formation or decomposition. This is so the rates of formation and rates of decomposition could cancel out.

Neil DSilva 1L
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### Re: If slope for ln(A) vs. t is negative is k?

Martha,

I don't think rates can be negative either. If k is always positive and concentrations are always positive, the rate is always positive as well. Correct me if I'm wrong, but even if the rate is for decomposition / consumption and has a negative sign, the rate is still actually positive, because the rate is given by (-1) x [(change in concentration) / (change in time)] where the change in concentration is given by the final concentration minus the initial concentration. Since the initial concentration is greater than the final concentration, the change is given with a negative value that is canceled out by the negative sign in front.

Chem_Mod
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### Re: If slope for ln(A) vs. t is negative is k?

Justin Le 2I wrote:Since the slope is equal to -k and the slope is -3, then k = 3. k is always positive.

Correct.
Slope = -k
-3 = -k
Therefore k = 3

Chem_Mod
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### Re: If slope for ln(A) vs. t is negative is k?

Equilibrium constants, rate constants, and the rate of a reaction are always positive numbers.

$rate = -\frac{d[R]}{dt}$
Since $\frac{d[R]}{dt}$ is negative due to decreasing reactant concentration the rate is positive.

Writing the rate with respect to the product gives,
$rate = \frac{d[P]}{dt}$ and this is also a positive value.

martha-1I
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### Re: If slope for ln(A) vs. t is negative is k?

Oh I understand. Thank you for the clarification!