Posted by mel on Tuesday, June 12, 2012 at 2:15pm.
The firstorder rate constant for photodissociation of A is 5.97 1/h. Calculate the time needed for the concentration of A to decrease to the following quantities.
a. 5% of its original value h
b. onethird of its initial concentration h

chemistry  DrBob222, Tuesday, June 12, 2012 at 3:20pm
k = 5.97 WHAT.
ln(No/N) = kt
ln(100/5) = 5.97*t
Solve for t.
I don't know the units (seconds, min, hours, days, years) because I can't tell from your value of 5.97 1/h.

chemistry  mel, Tuesday, June 12, 2012 at 3:37pm
5.97 1/hour

chemistry  DrBob222, Tuesday, June 12, 2012 at 3:44pm
Then the time is in hours.
b is worked the same way.

chemistry  mel, Tuesday, June 12, 2012 at 4:13pm
could yo explain part b?

chemistry  DrBob222, Tuesday, June 12, 2012 at 4:35pm
Same equation.
You can make up any number you want for No, then N is 1/3 that. I would pick a convenient number for No such as 300, then N = 100, and solve for t.
No could be 1 and N = 1/3 etc.

chemistry  mel, Tuesday, June 12, 2012 at 4:38pm
Thank You Very !
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