Question

Ethyl chloride vapor decomposes by the first-order reaction C2H5Cl→C2H4+HCl The activation energy is 249 kJ/mol and the frequency factor is 1.6×1014s−1.

Find the fraction of the ethyl chloride that decomposes in 19 minutes at this temperature.

Express your answer using one significant figure.

Answer 1

rate law for 1st order reactions: =ln[X] = -kt + ln[A]
k = rate constant = 7.7 x 10⁻⁵s⁻¹ [{  k = A*e^( -Eₐ/ (RT)) }
Eₐ = activation barrier = 249 kJ/ mole.
249 kJ/mole = 249000 J/mole
R = gas constant = 8.314 J/ mole-K
T = 710K
A = frequency factor = 1.6 x 10¹⁴ s⁻¹
k = A*e^( -Eₐ/ (RT))
k = (1.6 x 10¹⁴) *e^( -249000 / (8.314 * 710))
k = 7.7 x 10⁻⁵ ] now
t = time = 19min = 1140 seconds
[A] = initial concentration. We have to pick a value for this so let's pick 1.
[X] = concentration at time t
Plug into equation:
ln[X] = -(7.7 x 10⁻⁵s⁻¹)*(1140 seconds) + ln[1]
ln[X] = -(7.7 x 10⁻⁵s⁻¹)*(1140 seconds) + ln[1]
ln[X] = -.08778
[X] = e^(-.08778) = .916
the concentration that decomposed is 1 - .916 = .084 decomposed
The formula for percent decomposed is 100% x concentration decomposed/initial concentration
100% x .084/1 = 8.4% decomposed.