Solving Rate Law Problem: Find k for A--> Products

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SUMMARY

The discussion focuses on calculating the rate constant k for the second-order reaction A → Products, governed by the rate law Rate = k[A]^2. Given an initial concentration of A at 2.00 M and a reaction time of 1 hour for 60% of A to react, the correct value of k is determined to be 2.1 × 10^-4 M^-1 s^-1. Participants clarify the use of the second-order reaction equation 1/[A] = 1/[Ao] + kt to derive the rate constant accurately.

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  • Understanding of second-order reaction kinetics
  • Familiarity with the rate law and its application
  • Knowledge of concentration units (Molarity)
  • Basic algebra for manipulating equations
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  • Study the derivation of the second-order reaction equation 1/[A] = 1/[Ao] + kt
  • Learn how to apply the integrated rate laws for different reaction orders
  • Explore the impact of temperature on reaction rates and rate constants
  • Investigate experimental methods for measuring reaction rates
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Chemistry students, educators, and professionals involved in chemical kinetics and reaction mechanism analysis will benefit from this discussion.

firyace
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I have given this problem:

The reaction:
A--> Products

Follows the Rate Law, Rate = k[A]^2. When the initial concentration of A is 2.00M, it takes 1 hour for 60% of A to react at 25 celsius. What is the value of the rate constant k at this temp?

-I try to sub 2.00 into A, and (2.00 * .6)/3600 (in seconds) for rate, but I always get a different answer:bugeye:

-the answer is suppose to be 2.1*10^-4 M^-1 s^-1


Many thanks!
 
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Are you familiar with the equations for 0 order, 1st order and 2nd order reactions? It is possible to mathematically derive them, but it seems like you are usually given them in a data sheet.

The equation for a 2nd order reaction such as the one you have been given is

1/[A] = 1/[Ao] +kt

From there all you need to do is stick in your numbers. You have been given all of the data you need to put numbers in for the initial and final concentration, and are given a time. Got it?
 

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