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Topic: Chemical Kinetics and decomposition  (Read 1997 times)

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Offline motherofcorgs

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Chemical Kinetics and decomposition
« on: September 08, 2018, 06:43:52 PM »
Hi everyone. I'm having a hard time understanding how to work this problem.

In a first order decomposition, the constant is 0.00737 sec-1. What percentage of the compound has decomposed after 93.4 seconds?

I tried to work it out like this:

ln(100)-ln(x)=0.00737 ·93.4 but my answer isn't correct. It says the answer should be 50. Will someone please explain how to do this for my next homework attempt?

The other question I have is this.
In a first order decomposition, how long does it take (in seconds) until 69% of the compound is left given that k is 2.26 min-1 (give answer to 1 decimal place)?

I tried to work it out like this:

ln(100)-ln(69)=135.6 ·t. I'm getting 0.0027 but it says the correct answer is 9.9. What am I doing wrong?

Offline mjc123

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Re: Chemical Kinetics and decomposition
« Reply #1 on: September 10, 2018, 04:58:54 AM »
Part i: Your formula looks correct. Did you just make an error in calculation?
Part ii: If your time constant is 2.26 min-1, how do you convert that into a time constant in sec-1? (BTW You don't need to; why not just do the calculation in minutes and convert to seconds at the end?)

Offline Babcock_Hall

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Re: Chemical Kinetics and decomposition
« Reply #2 on: September 10, 2018, 10:27:11 AM »
With respect to part one, 50% means that 0.50 has decomposes and 0.50 remains.  I trust that you correctly converted into the percent scale.

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