Ethyl chloride vapor decomposes by the first-order reaction: C_{2}H_{5}{Cl}\r

eiraszero11cu 2021-12-17 Answered
Ethyl chloride vapor decomposes by the first-order reaction: C2H5{Cl}C2H4+HCl. The activation energy is 249kjmol, and the frequency factor is 1.6×1014s1. Find the value of the rate constant at 710 K. What fraction of the ethyl chloride decomposes in 15 minutes at this temperature? Find the temperature at which the rate of the reaction would be twice as fast.
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Expert Answer

kalfswors0m
Answered 2021-12-18 Author has 24 answers

Rate constant at 710 K can be calculated using the one-point Arrhenius equation.
k=Ae(EaRT)
Given are A=1.6×1014s1
Ea=249kJmolor249000Jmol
R=8.314JmolK
T=710K
Plugging in
k=(1.6×1014{s}1)e(d{249000Jmol}8.314{J}{mol}{K}{710K})=7.67×105s1

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Maria Lopez
Answered 2021-12-19 Author has 32 answers

Use the first-order integrated rate law to calculate the fraction of ethyl chloride that decomposed after 15 mins.
ln([A]t[A]o)=kt
Isolating the ratio of the remaining ethyl chloride [A]t[A]o
[A]t[A]o=ekt
plugging in k=7.67×105{s}1 and t=15mins60seconds1min=900seconds.
[A]t[A]o=e(7.67×105s1)(900 seconds)=
The fraction of remaining ethyl chloride is 0.9333. The fraction of ethyl chloride that decomposed is 10.9333=0.067.

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Don Sumner
Answered 2021-12-27 Author has 35 answers

The temperature where the rate of of reaction is twice than the rate of reaction at 710 K can be calculated using the two-point Arrhenius equation.
ln(k2k1)=EaR(1T11T2)
If we assume k1 is the rate constant at T1=710K then k2  is the rate constant at T2=? where k2=2k1.
R=8.314JmolK
Isolate T2
ln(k2k1)=EaR(1T11T2)
ln(k2k1)REa=(1T11T2)
1T2=1T1ln(k2k1)REa
T2=11T1ln(k2k1)REa
plugging in
T2=11710Kln(2k1k1)8.314JmolK249000J/mol
T2=721.86K
Therefore, T2=720K

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