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

[Nicole Leppi 2K]

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]

that's what I thought, thanks!

[Justin Le 2I]

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

[martha-1I]

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]

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]

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]

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


Since is negative due to decreasing reactant concentration the rate is positive.

Writing the rate with respect to the product gives,
and this is also a positive value.

[martha-1I]

Oh I understand. Thank you for the clarification!