Cyclopropane, C3H6, is converted to its isomer propylene, CH2=CHCH3, when heated. The rate law is first order in cyclopropane, and the rate constant is 6.0 10^-4 /s at 500°C. If the initial concentration of cyclopropane is 0.0238 mol/L, what is the concentration after 719 s?
ln(No/N) = kt
ln(0.0238/N) = 6.0E-4*719
N = ?
To determine the concentration of cyclopropane after 719 seconds, we can use the first-order rate equation:
ln[Cyclopropane]t = -kt + ln[Cyclopropane]0
where [Cyclopropane]t is the concentration at time t, k is the rate constant, and [Cyclopropane]0 is the initial concentration.
Given:
Rate constant (k) = 6.0 x 10^-4 /s
Initial concentration ([Cyclopropane]0) = 0.0238 mol/L
Time (t) = 719 s
Plugging in the values, we get:
ln[Cyclopropane]719 = -6.0 x 10^-4 /s * 719 s + ln(0.0238 mol/L)
Now we can calculate the concentration of cyclopropane after 719 seconds by solving for [Cyclopropane]719:
ln[Cyclopropane]719 = -0.4314 + ln(0.0238 mol/L)
Using the natural logarithm of both sides, we can find:
[Cyclopropane]719 = e^(ln[Cyclopropane]719)
[Cyclopropane]719 = e^(-0.4314 + ln(0.0238 mol/L))
Calculating the value:
[Cyclopropane]719 ≈ 0.0191 mol/L
Therefore, the concentration of cyclopropane after 719 seconds is approximately 0.0191 mol/L.
To find the concentration of cyclopropane after a certain time, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt,
where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
In this case, the concentration of cyclopropane ([A]0) is given as 0.0238 mol/L, the time (t) is given as 719 s, and the rate constant (k) is given as 6.0 * 10^-4 /s.
Plugging these values into the equation, we have:
ln([A]t/0.0238) = -(6.0 * 10^-4 /s) * 719 s.
To solve for [A]t, we need to isolate it in the equation. Here are the steps to solve it:
1. Multiply both sides of the equation by 0.0238:
ln([A]t/0.0238) * 0.0238 = -(6.0 * 10^-4 /s) * 719 s * 0.0238.
2. Simplify the right side:
ln([A]t/0.0238) * 0.0238 = -(0.9562 * 10^-3) * 719.
3. Take the exponential of both sides of the equation to remove the natural logarithm:
e^(ln([A]t/0.0238) * 0.0238) = e^(-(0.9562 * 10^-3) * 719).
4. Simplify the left side:
[A]t/0.0238 = e^(-(0.9562 * 10^-3) * 719).
5. Multiply both sides of the equation by 0.0238 to solve for [A]t:
[A]t = 0.0238 * e^(-(0.9562 * 10^-3) * 719).
Plugging the values into a calculator, we find:
[A]t ≈ 0.00623 mol/L.
Therefore, the concentration of cyclopropane after 719 seconds is approximately 0.00623 mol/L.