For the reaction, A = B + C, the rate law is kA]. If it takes 80.0 seconds for 70.0% of a 10.0 gram sample of A to be transformed into products, what is the value of the rate constant?
a. 0.00450 s-1 b. 0.0290 s 1 c. 0.00530 s-1 d. 0.0150 s-1 e. 5.40 s-1
chemistry - DrBob222, Friday, February 17, 2012 at 10:06pm
rate = k(A) is first order.
Use ln(No/N) = kt. Substitute and solve for k.
B?
I get the answer .01505 is that correct?
Surely you didn't try this. Plug in those numbers and try again.
No = 10
N = ? if 70% has been transformed.
k ?
t = 80 sec.
yes
To find the value of the rate constant (k), we can use the given information and the first-order rate equation, ln(No/N) = kt.
First, let's define our variables:
No = initial amount of substance A
N = amount of substance A at a given time t
t = time taken for transformation (80.0 seconds)
k = rate constant (what we need to find)
We are given that 70.0% of the 10.0 gram sample of A has been transformed into products after 80.0 seconds. So, N = 0.3 * No (since 70% of the initial amount remains).
Substituting these values into the rate equation, we have:
ln(No/N) = kt
ln(No/(0.3 * No)) = k * 80.0
Simplifying the equation further:
ln(1/0.3) = 80.0k
ln(10/3) = 80.0k
Now, we can solve for k by dividing both sides of the equation by 80.0:
k = ln(10/3)/80.0
Using a calculator, we find that k is approximately 0.00530 s-1.
Therefore, the correct answer is c. 0.00530 s-1.