Cyclpropane is converted to its isomer propylene, when heated. The rate law is first order in cyclopropane and the rate constant is 6.0X10^-4/s at 500 degrees C. If the intial concentration of cyclopropane is 0.0226 mol/L, what is the concentration after 899s?
To determine the concentration of cyclopropane after 899 seconds, we can use the first-order rate law equation:
ln(Ct/C0) = -kt
Where:
Ct is the concentration at time t
C0 is the initial concentration
k is the rate constant
t is the time elapsed
Rearranging the equation, we have:
Ct = C0 * e^(-kt)
Now we can plug in the values provided in the problem statement:
C0 = 0.0226 mol/L (initial concentration)
k = 6.0 × 10^-4/s (rate constant)
t = 899 s (time elapsed)
Plugging in these values into the equation, we have:
Ct = 0.0226 mol/L * e^(-6.0 × 10^-4/s * 899 s)
Simplifying this equation, we can calculate the concentration:
Ct = 0.0226 mol/L * e^(-0.5394)
To find the concentration of cyclopropane after 899 seconds, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt
Where:
[A]t = concentration of cyclopropane after time t
[A]0 = initial concentration of cyclopropane
k = rate constant
t = time
Rearranging the equation, we have:
[A]t = [A]0 * e^(-kt)
Now let's plug in the given values into the equation:
[A]0 = 0.0226 mol/L (initial concentration of cyclopropane)
k = 6.0 × 10^(-4) /s (rate constant)
t = 899 s (time)
[A]t = 0.0226 mol/L * e^(-6.0 × 10^(-4)/s * 899 s)
Using the exponential function, we can approximate e^(-6.0 × 10^(-4)/s * 899 s) as e^(-0.5394) which is approximately 0.5832.
[A]t = 0.0226 mol/L * 0.5832
Now we can calculate the concentration of cyclopropane after 899 seconds:
[A]t = 0.0132 mol/L
Therefore, the concentration of cyclopropane after 899 seconds is approximately 0.0132 mol/L.