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
rate = k(A) is first order.
Use ln(No/N) = kt. Substitute and solve for k.
To solve this problem, we need to use the integrated rate law for a first-order reaction.
The integrated rate law for a first-order reaction can be written as:
ln(A₀ / A) = kt
Where:
- A₀ is the initial concentration of A
- A is the concentration of A at a given time
- k is the rate constant for the reaction
- t is the time
In this case, we are given that it takes 80.0 seconds for 70.0% of a 10.0 gram sample of A to be transformed into products. This means that the concentration of A is 30.0% of the initial concentration.
Using this information, we can rewrite the integrated rate law as:
ln(1 / 0.3) = k(80.0 s)
Simplifying, we have:
ln(10/3) = 80k
To solve for k, we rearrange the equation:
k = ln(10/3) / 80
Calculating this expression gives us:
k ≈ 0.004382 s⁻¹
Now, we need to choose the closest value from the options provided. The closest value is 0.00450 s⁻¹, which corresponds to option a.
So, the correct answer is:
a. 0.00450 s⁻¹