Posted by **mel** on Tuesday, June 12, 2012 at 2:15pm.

The first-order 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. one-third 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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