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
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.
5.97 1/hour
Then the time is in hours.
b is worked the same way.
could yo explain part b?
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.
Thank You Very !
To calculate the time needed for the concentration of A to decrease to a certain value, we can use the first-order rate equation:
A(t) = A0 * e^(-kt)
where A(t) is the concentration of A at time t, A0 is the initial concentration of A, k is the first-order rate constant, and e is the base of the natural logarithm.
a. To calculate the time needed for the concentration of A to decrease to 5% of its original value, we need to solve for t when A(t) is equal to 0.05 * A0:
0.05 * A0 = A0 * e^(-kt)
Dividing both sides of the equation by A0, we get:
0.05 = e^(-kt)
To solve for t, we can take the natural logarithm of both sides of the equation:
ln(0.05) = -kt
Now we can plug in the values and solve for t:
t = ln(0.05) / (-k)
Substituting the value of k given in the question (k = 5.97 / h):
t = ln(0.05) / (-5.97 / h)
Calculating this expression will give you the time needed for the concentration of A to decrease to 5% of its original value.
b. To calculate the time needed for the concentration of A to decrease to one-third of its initial concentration, we repeat the same steps as in part (a), but instead, we set A(t) to 1/3 * A0 and solve for t:
1/3 * A0 = A0 * e^(-kt)
Dividing both sides of the equation by A0, we get:
1/3 = e^(-kt)
Taking the natural logarithm of both sides gives:
ln(1/3) = -kt
Now we can substitute the value of k and solve for t:
t = ln(1/3) / (-k)
Substituting the given value of k (k = 5.97 / h) will give you the time needed for the concentration of A to decrease to one-third of its initial concentration.