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

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.

Answer 1

1) For a first order reaction,

ln[A] = ln[Ao] - kt

[A] = fraction of ethyl chloride remaining

[Ao] = initial concentration of ethyl chloride

k = 1.4 x 10^-4 s-1

t = 19 x 60 = 1140 s

Feed values,

ln([A]/[Ao]) = -1.4 x 10^-4 x 1140

[A]/[Ao] = 0.852

fraction of ethyl alcohol decomposed = 1 - 0.852 = 0.148

2) Temperature at which rate would be twice fast

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

k2/k1 = 2

Ea = 249 kJ/mol

R = gas constant

T1 = 720 K

T2 = ?

we get,

ln(2) = 249000/8.314 (1/720 - 1/T2)

T2 = 732.20 K

Is the temperature required