First Order Reaction Rate

[Drew Myers 4G]

How do you get to ln[R] = -kt + ln[R]initial from -(1/a)(d[R]/dt) = k[R]^n (where n = 1)?

[Suraj Doshi 2G]

This equation comes from separating the variables in the differential rate law and then taking its integral.

[romina_4C]

Since in a first order reaction, the coefficient of the reactants is one, and "a" represents the coefficient of the reactant, thus "a" = 1. Thus, the equation becomes -d[A]/dt = k[A]. Normally on the right side of the equation [A] would be raised to the "n" power, but since the coefficient of A is one, then it simply becomes [A]. After that, we multiply around to get everything with A in it on one side of the equation and everything with "t" on it to the other side of the equation. We do this so that we can integrate. Thus, we obtain d[A]/[A] = -kdt (also multiply the negative over). We can now integrate to get ln[A] = -kt + C. Setting t = 0, C = ln[A initial] (consider this in terms of the reaction, at time = 0, there is the initial amount of the reactant left). Therefore, the final equation is ln[A] = -kt + ln[A initial].

[Adriana_4F]

Does anyone know if this will be on Test 2?

[Louise Lin 2B]

Does anyone know if this will be on Test 2?

I believe that test 2 only goes up till electrochemistry, and rate laws are part of kinetics, so I don't think rate laws will be on test 2.