Consider the reaction A ==> B. If the reaction is first-order in A with the initial concentration A being 6.048M, what is the concentration of A after 1.986 seconds if the rate constant is 0.483s-1?
Error tolerance: ±0.5%
Can someone explain how to do this? Thanks
ln(Ao/A) = kt
You know Ao, solve for A, you know k and you know t.
To solve this problem, you can use the first-order reaction equation: ln(Ao/A) = kt, where Ao is the initial concentration of A, A is the concentration of A at time t, k is the rate constant, and t is the time elapsed.
In this case, you are given Ao = 6.048M, k = 0.483s^-1, and t = 1.986 seconds. You need to calculate the concentration of A after 1.986 seconds.
To start, rearrange the equation to solve for A:
ln(Ao/A) = kt
Taking the antilog of both sides, we have:
Ao/A = e^(kt)
Next, substitute the given values into the equation:
Ao/A = e^(0.483s^-1 * 1.986s)
Now, calculate the concentration of A by solving for A:
A = Ao / e^(0.483s^-1 * 1.986s)
A = 6.048M / e^(0.961s^-1)
Using the value of e (approximately 2.718), we can evaluate the expression:
A = 6.048M / (2.718^(0.961))
A = 6.048M / 2.614
A ≈ 2.314M
Therefore, the concentration of A after 1.986 seconds is approximately 2.314M, given the rate constant and initial concentration provided.
Please note that the final answer is an approximation and may have slight variation within the error tolerance of ±0.5%.