A certain substance undergoes decomposition by a first-order reaction which has equal to 30.0 minutes. If the initial concentration of the substance is 1.28 mol/liter, what will its concentration be after 2.50 hours?
1. 0.36 M 2. 0.040 M 3. -0.32 M 4. 0.020 M 5. 0.16 M
6. None of the previous answers
My Work:
ln(2)/30minutes=k=.023104
First order reaction:
ln[At]=ln[Ao]-kt
Change t to minutes=150minutes
ln[At]=ln[1.28mol/l]-(.023104mins-1)*(150minutes)
At=.040M, which would be 2 if I did it right.
If I did do it right, do you think my teacher is trying to be tricky with significant figures because .0400 would be 3 significant figures, which is how many each of the numbers has in the problem? Thank you so much!!
The first part of your question I can't answer because you appear to have omitted a word in "...which has equal to ...."
The second part of your question about tricky profs....probably so; however, I don't see any answers with any more than two s.f.
Oh sorry, 30 minutes is supposed to be the half life, so it should say first order reaction which has a half life (t1/2) equal to 30 minutes. I know, do you have any advice on what you'd do for the significant figures part? Would you go with the answer on there or do none of the answers? I'm just stressed about it. Thank you!
I think you are stressed out over nothing. The answer I get is 0.04002945 which to two s.f. (limited I think by the 2 places in 30 minutes) is 0.040. I would go with answer 2.
Your calculations are correct. The concentration of the substance after 2.50 hours is indeed 0.040 M.
Regarding significant figures, it's always a good idea to follow the rules of significant figures when reporting your final answer. In this case, since the initial concentration was given to three significant figures (1.28 M), you should also report the final concentration to three significant figures, which would be 0.040 M.
It is possible that your teacher is testing your understanding of significant figures, so be sure to follow the correct rules.