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### Course: AP®︎/College Chemistry>Unit 5

Lesson 3: Concentration changes over time

# Worked example: Using the first-order integrated rate law and half-life equations

In this video, we'll use the first-order integrated rate law to calculate the concentration of a reactant after a given amount of time. We'll also calculate the amount of time it takes for the concentration to decrease to a certain value. Finally, we'll use the first-order half-life equation to calculate the half-life of the reaction. Created by Jay.

## Want to join the conversation?

• Sir ,isn't there any other method to find the concentration other than using ln and exponential? Because during our calculations we cannot use a calculator which is pretty tough.
• If you are referring to the MCAT exam, from what I have heard, the MCAT tests more conceptual knowledge about rate laws and simpler math, such as finding rate laws rather than complex problems like this. I think it is important to understand the main ideas behind this problem, such as half-life, but I don't think we will be expected to do this much math.
• is the half life always t1/2 = .693/K ?
• Only if the reaction is a first order reaction.
• At , why don't you divide the two ln's? I thought you were supposed to divide when you subtract.
• well, he could have written it as 'ln(0.01/0.05)' but his method is more straightforward
(1 vote)
• how would you find the starting concentration at time= zero if you were given a set of data (time vs concentration)?
(1 vote)
• I would plot ln(c) vs. t and extrapolate the straight line back to t=0.
• How much math knowledge is needed for AP Chemistry? Do I need to know Pre-Calc?
• At a minimum, I would say algebra since it can get you through a majority of chemistry problems. But the more math you know, the better your problem solving capabilities.
(1 vote)
• is it that the average rate of a first-order reaction always gets slower with time?
• In the worked example video: Are significant figures of any importance?
(1 vote)
• Yes, they're important. You can't do science, at least professionally, without using correct significant figures.
• Shouldn't we round 2402 seconds to 2400 seconds, and 1034 seconds to 1000 seconds, because we only have two sig-figs in the rate constant? ,
(1 vote)
• You're correct, both those answers should be correctly reported with only two sig figs because of the rate constant. However writing 2400 and 1000 seconds can be misconstrued as having differnt amounts of sig figs and are considered ambigious answers. For example one person could argue that 2400 has two sig figs (the 2 & 4) while another person could argue that all four digits are significant.

To avoid this issue we should write them in scientific notation. So 2400 s becomes 2.4 x 10^(3) s, and 1000 becomes 1.0 x 10^(3) s. And now everyone is clear that both numbers have two sig figs each.

Hope that helps.
• do you have to convert the rate constant to seconds when determining the half live, if the rate constant is given in hours? or would you find the half live using the hours k and then convert the units to seconds?
(1 vote)
• What happened to the units? I imagine that the logarithm only makes sense if the quantity is dimensionless, but I don't understand how we just make something unitless.
(1 vote)
• Taking a logarithm of something doesn’t make it unitless. Taking the ln([A]) still results in a unit of molarity (M) since that is the unit of [A]. Only canceling units through division makes numbers dimensionless (unitless).

Jay doesn’t include the units throughout the problem, only the numerical values since it makes the calculations more compact. However, he does include units for the answers which is where they are truly needed.

If you follow the first-order integrated rate law using units, it works out.
Ln([A]t) = -kt + ln([A]0)
M = -(1/s)(s) + M
M = M

Hope that helps.
(1 vote)