Question

Ethyl chloride vapor decomposes by the first-order reaction:

Answer 1

Explanation:

(a)  The given data is as follows.

            Frequency factor A = $1.60 \times 10^{14} s^{-1}$

           Activation energy ($E_{a}$) = 249 kJ/mol

           Gas constant = 8.314 J/mol K

            Temperature (T) = 876 K

Hence, calculate the specific rate constant as follows.

                  K = $Ae^{\frac{-E_{a}}{RT}}$

                      = $1.60 \times 10^{14} s^{-1} e^{\frac{249000 J/mol}{8.314 J/mol K \times 876 K}}$

                      = $1.60 \times 10^{14} s^{-1} \times 1.034$

                      = $1.128 \times 10^{29}$

Hence, the value of specific rate constant K is $1.6544 \times 10^{14}$.

(b)  It is given that time is 20 minutes. As there are 60 seconds present in 1 min. Therefore, seconds present in 20 minutes are as follows.

            $20 min \times \frac{60 sec}{1 min}$

            = 1200 sec

Hence, for the first order reaction the fraction remaining is as follows.

                $\frac{C}{C_{o}} = e^{-kt}$

                              = $e^{-1.6544 \times 10^{14} \times 1200 sec}$

                              = 0

Hence, fraction decomposed = 1 - 0 = 1

Therefore, the fraction of the ethyl chloride that decomposes in 20 minutes at this temperature is 1.