The previous exercise shows how the first-order integrated rate law is derived from the first-order differential rate

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The previous exercise shows how the first-order integrated rate law is derived from the first-order differential rate law. Begin with the second-order differential rate law and derive the second-order integrated rate law.


Previous Exercise

The first-order integrated rate law for the reaction A → products is derived from the rate law using calculus:Rate = k[A] (first-order rate law) d[A] Rate = dt d[A] dt -k[A]

The equation just given is a first-order, separable differential equation that can be solved by separating the variables and integrating:d[A] [A] IA]d[A] [A]o [A] -kdt -S kdt

In the integral just given, [A]0 is the initial concentration of A.

We then evaluate the integral:[In[A]] = k[t] In[A] In[A]o -kt In[A] = kt + In[A]o (integrated rate law)

Use a procedure similar to the one just shown to derive an integrated rate law for a reaction A → products, which is onehalf order in the concentration of A (that is, Rate = k[A]1/2).

Use the result from part a to derive an expression for the halflife of a one-half-order reaction.

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