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 solve this problem, we need to use the first-order rate law equation, which is as follows:
Rate = k * [Cyclopropane]
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [Cyclopropane] is the concentration of cyclopropane
Given:
- k = 6.0 × 10^-4/s (rate constant)
- [Cyclopropane] = 0.0226 mol/L (initial concentration)
- t = 899s (time)
We can rearrange the rate law equation to solve for [Cyclopropane]:
[Cyclopropane] = Rate / k
To find the rate, we can use the rate constant and the initial concentration:
Rate = k * [Cyclopropane] = (6.0 × 10^-4/s) * (0.0226 mol/L)
Now, we can substitute the values and calculate the rate:
Rate = (6.0 × 10^-4/s) * (0.0226 mol/L)
= 1.356 × 10^-5 mol/(L*s)
Now, we can substitute the rate into the equation to find the concentration after 899s:
[Cyclopropane] = Rate / k
= (1.356 × 10^-5 mol/(L*s)) / (6.0 × 10^-4/s)
= 0.0226 mol/L
Therefore, the concentration of cyclopropane after 899s is 0.0226 mol/L.