Find the temperature at which the rate of the reaction would be twice as fast.Et

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
C2H5ClC2H4+HCl
The activation energy is 249 kJ/mol and the frequency factor is 1.6×1014s1.

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.

Explanation / Answer

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

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