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

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.

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
The value of the specific rate constant at 720 K is k= 1.4x10^-4
s^-1
1) Find the fraction of the ethyl chloride that decomposes in 19
minutes at this temperature.
2) Find the temperature at which the rate of the reaction would
be twice as fast.

Find the temperature at which the rate of the reaction would be
twice as fast.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.
Part A Find the value of the specific rate constant at 700 K
.
Part B
Find the fraction of the ethyl chloride that decomposes in 20
minutes at this
temperature.[C2H5Cl]0−[C2H5Cl]t[C2H5Cl]0[C2H5Cl]0−[C2H5Cl]t[C2H5Cl]0
Part C= Find the temperature at which the rate of the reaction
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