CAS CH 110 Lecture Notes - Lecture 29: Rate Equation

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25 Apr 2017

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Summary: using initial rates experimental data we can find a rate law expression, rate = k [a]a[b]b the orders a, b, and constant, k. Now, we extend our initial rates ideas into a way to predict concentrations; this will require calculus. Integrating the three rate laws will give us future concentrations. Start by using our first order rate expression, rate = k[a]. Our goal is to express the rate as [a] changes over time. Rate of disappearance of a: - d[a] /dt = k[a] Divide both sides by [a] to keep the [a]"s together and move the dt so we can integrate the expression. Now we integrate for concentration and for time. The half-life, t1/2 , is the time needed for its concentration to fall to onehalf of its initial value. The higher the value of k, the more rapid the consump-tion of a reactant.

Related Questions

114. The A — first-order integrated rate law for the reaction products is derived from the rate law using calculus: Rate = k[A] (first-order rate law) d[A] Ratedt [A] =- [A] The equation just given is a first-order, separable differential equation that can be solved by separating the variables and inte- grating: d[A] = -kdt TAJ PINA[A] - kat JA), [A] kat In the integral just given, [A]o is the initial concentration of A. We then evaluate the integral: [In[A]] A) = -k[t]6 In[A] - In[A]o = -kt In[A] = -kt + In[A]o (integrated rate law) a. Use a procedure similar to the one just shown to derive an in- tegrated rate law for a reaction A — products, which is one- half order in the concentration of A (that is, Rate = k[A]"). b. Use the result from part a to derive an expression for the half- life of a one-half-order reaction.

if a reactant is used up according to a first order rate law, and the initial concentration of the reactant is 3.2 mol L-1 , what is the concentration of the r reactant after two half-lives have passed and after six half-lives have passed?

Different Q, For each of the following rate laws, describe what would happen to the reaction rate if the concentration of reactant A was triple and the concentration of reactant B is halved. Rate=k[A] [B] Rate= k [A]2 [B] Rate=k [A]2 [B]2 rate= k [A] [B]3

goldotter472

What rate law and integrated rate lets you calculate the time required to let your [CV⁺] change to half of its value at t=0? Second order rate law and its integrated rate is useful to calculate [CV⁺] to half life at t =0.

We want to find the rate law that describes how reactant concentrations affect the rate of the reaction shown by: CV⁺ + OH^- --------------- CVOH

The general form of the rate law we will find is: rate = k[CV⁺]^x [OH^-]

where x, y are the reaction orders with respect to reactanst, x+y is the overall reaction order and k is the rate constant.

so far, I have found rate = 2.0 * 10^-6 and k=1.25.

Now, I need to calculate the time needed for the concentration to be halved.

CAS CH 110 Lecture Notes - Lecture 29: Rate Equation

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