Question

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 would be twice as fast.

Answer 1

A) we have formula k= A exp ( -Ea /RT)

k = 1.6 x 10^ 14 exp ( - 249000 /8.314 x700)

    = 4.196 x 10^ -5 s^-1

B) we have formula for 1st order kinetcis     t = ( 2.303/k) log ( a/a-x)    where t = 20 min = 20 x 60 = 1200 s

1200 = ( 2.303/4.196 x 10^ -5) log ( a/a-x)

fraction = (a-x) / a = 0.95

a = initail amount = 1 then a-x = 0.95   = amount left

fraction decomposed = x = a- ( a-x) = 1-0.95 = 0.5 , hence 0.5 fraction dcomposed.

C) we have formula ln ( k2/k1) = (Ea/R) ( 1/T1 -1/T2)

ln ( 2 ) = ( 249000/8.314) ( 1/700 - 1/T2)

T2 = 711.5 K