September 19, 2021, 04:37:00 PM
Forum Rules: Read This Before Posting


Topic: Half-life  (Read 485 times)

0 Members and 1 Guest are viewing this topic.

Offline ist2601

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Half-life
« on: March 26, 2021, 08:01:21 AM »
Ethyl ethanoate undergoes irreversible second order reaction hydrolysis reaction with sodium hydroxide in mixed water-methanol solution. Running the reaction at 303K the order of this reaction with respect to ethyl ethanoate is equal to one and half-life of this reactant is equal to 1800s. Taken the reactants of this reaction in stoichometric proportions, calculate time it took for 10% of ethyl ethanoate to hydrolyse under those conditions.

1. My first question to this task is if I correct recognize the type of problem and I started from correct formulas.

2. My second question is can I assume that if reactant was taken in stoichometric proporction it is 1 mole of ethyl ethanoate and 1 mole of sodium hydroxide ?

I attach my notes in the picture

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 1946
  • Mole Snacks: +281/-12
Re: Half-life
« Reply #1 on: March 27, 2021, 12:11:43 PM »
The approach is correct. However, I'd just make a couple of points.
1. C°A is not 1 mole. What are the units of C?
2. Stoichiometric proportions doesn't mean that the starting concentrations are both 1 M, but simply that they are both the same. Note that for a second-order reaction, unlike a first-order reaction, the half-life is not constant but depends on the initial concentration - which you don't know. However, the ratio of the half-life to the 0.1th life is independent of initial concentration. You might like to try and verify this algebraically. (Calculate k as a function of C° in the first step, then calculate t0.1, and see if C° cancels out.)

Sponsored Links