The rate constant for the equation, 2C2F4 → C4F8, is 0.05 M^−1· s^−1. We start with 0.111 mol C2F4 in a 3.00-liter container, with no C4F8 initially present. What will be the concentration of C2F4 after 1.00 hour? Answer in units of M.
To find the concentration of C2F4 after 1.00 hour, we can use the first-order kinetic equation:
[A]t = [A]0 * e^(-kt)
Where:
[A]t = concentration of reactant A at time t
[A]0 = initial concentration of reactant A
k = rate constant
t = time
In this case, the reactant A is C2F4, the initial concentration [A]0 is 0.111 mol/L, the rate constant k is 0.05 M^−1· s^−1, and the time t is 1.00 hour.
First, we need to convert the time to seconds. Since there are 3600 seconds in 1 hour, 1.00 hour is equal to 3600 seconds.
Now, we can calculate:
[A]t = 0.111 mol/L * e^(-0.05 M^−1· s^−1 * 3600 s)
Let's break down the calculation into steps:
1. Multiply the rate constant by time:
(-0.05 M^−1· s^−1) * (3600 s) = -180 M^−1
2. Plug the value into the equation:
[A]t = 0.111 mol/L * e^(-180 M^−1)
3. Use a scientific calculator to evaluate the exponential term:
e^(-180 M^−1) = 9.8 x 10^(-79)
4. Multiply the calculated exponential term by the initial concentration:
[A]t = 0.111 mol/L * (9.8 x 10^(-79))
5. Calculate the final answer:
[A]t ≈ 1.1 x 10^(-80) M
Therefore, the concentration of C2F4 after 1.00 hour is approximately 1.1 x 10^(-80) M.