The unit of rate constant of a first order reaction is

The unit of the rate constant of a first-order reaction is s - 1 Step 1: Write differential rate law: • According to the differential rate law • For the reaction A → B • Rate = - d A d t Step 2: Write the rate law: • According to rate law • For the reaction A → B • Rate ∝ A x where x is the order of reactant A • Rate = K A x where K is the rate constant of reaction and [ A ] is the concentration of reactant Step 3: Finding the unit for rate of the reacton: A reaction in which the rate depends on the concentration of only one reactant is called a first-order reaction. • Therefore rate of reaction = - dA dt = K A x A x • Unit or rate of reaction = mol L - 1 s - 1 • For a first-order reaction:- • Rate = - d A d t = K A 1 • Unit of the rate of reaction = = mol L - 1 s - 1 • Unit of concentration = mol L - 1 • K = Rate A 1 = mol L - 1 s - 1 mol L - 1 K = s - 1 • Therefore unit of the rate constant of a first-order reaction is s - 1 .

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The integrated rate law for the first-order reaction A → products is ln[A]_ t = - kt + ln[A]_0. Because this equation has the form y = mx + b, a plot of the natural log of [A] as a function of time yields a straight line. The rate constant for the reaction can be determined from the slope of the line, which is equal to - k. Created by Jay. This is grade-12/college-level but if you're curious I will show you below. So for a first order reaction -- we have the reaction equals the rate constant times the concentration of the (only) reactant --> R = k[A] 1. Then we choose to re-write R as -Δ[A]/Δt and we get -Δ[A]/Δt = k[A] 2. Then we bring -Δt to the right side Δ[A] = -k[A]Δt 3. Then we bring [A] to the left side Δ[A]/[A] = -kΔt 4. Then we integrate (the left side with respect to A and the right side with respect to t) ∫Δ[A] 1/[A] = -k∫Δt Ln[A] = -kt 5. Then we evaluate both integrals from 0 to t Ln[A]ₜ - Ln[A]₀ = -kt-(-k0) 6. Then we bring Ln[A]₀ to the right Ln[A]ₜ = -kt -0 + Ln[A]₀ 7. Finally, we have our answer: Ln[A]ₜ = -kt + Ln[A]₀ 8.Notes: i. For the connection to y=mx+b The natural log of the concentration of A at a given time t --> Ln[A]ₜ is basically Y, and is equal to the natural log of the initial concentration of A --> Ln[A]₀ which is basically b, minus the rate constant -->k (basically m, aka the slope of the line) multiplied by time (basically x). So we get a linear graph of the form Y=mx+b ii. The reason the it is negative at the beginning -Δ[A]/Δt and at -kt ...

I have a little confusion about first order reactions that produce products being dependent upon the concentration of the reactant, (i.e. if you double the reactant in a first order reaction, you double the amount of product produced), while the half life decay of a first order reaction that produces a product (i.e. half the initial concentration of the reactant) is not dependent upon the initial concentration. Could you please explain these two differences when you get a chance? By the way, these videos have been extremely helpful and I appreciate all the hard work you and Khan Academy have put into making these materials available to anyone, anytime, anyplace. In earlier videos we see the rate law for a first-order reaction R=k[A], where [A] is the concentration of the reactant. If we were to increase or decrease this value, we see that R (the rate of the reaction) would increase or decrease as well. When dealing with half-life, however, we are working with k (the rate constant). While the rate of reaction is measured in units molar/second, a rate constant for a first-order reaction is 1/(second). If you graph half life data you get an exponential decay curve. It’s kind of the definition of it. If you graph something that starts at 100 and decays by half every 1 minute, 50 by minute 2, 25 by 3, 12.5 by 4, 6.25 by 5 etc. you’ll see. Go back to the previous video and look at the label on the y axis, then compare with the y axis on this one. A quick method for working out t...

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. 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. 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. We can, but only for a first order reaction. Each order has its own half-life equation. Zeroth order: ([A0]/2k) First order: (ln(2)/k) Second order: (1/k[A0]) So the zeroth ...

In many textbooks, it is written that: The value of rate constant depends on the nature of the reactants, temperature and catalyst. It is independent of the concentration of the reactants. However, the unit of the rate constant is $$\left(\frac$ for all other reactions. Does this contradict the statement given in books? $\begingroup$ @EashaanGodbole I think units matter. However, you can't tell from the units what a quantity depends on. Let's say I drive with a constant speed of 100 km/h on the freeway. Just because the units have the length-unit km in them does not mean my speed depends on how far I've already driven (or how long, for that matter). The idea of constant speed is that the ratio of distance and time is constant, even as time passes and you move along. $\endgroup$ As an example if the reaction is $\ce.$ It is true that the numerical value of the rate constant usually depends of several things such as temperature, but the units remain the same for each type of rate constant, 0 th, 1 st, 2 nd, 3 rd simply because the rate equation units have to balance. Thanks for contributing an answer to Chemistry Stack Exchange! • Please be sure to answer the question. Provide details and share your research! But avoid … • Asking for help, clarification, or responding to other answers. • Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. To learn more, see our