For the first order thermal decomposition reaction, the following data were obtained:
C$_2$H$_5$Cl$_g$ $\xrightarrow{h\nu}$ C$_2$H$_4$$_{(g)}$ + HCl $_{(g)}$

Time (sec)	Total pressure (atm)
0	0.30
300	0.50

Calculate the rate constant. (Given: log 2 = 0.301, log 3 = 0.4771, log4 = 0.6021)

Hint: We are given with the values of pressure, at the different time. We will calculate the total pressure at time t. The reaction is given in which there is decomposition of ethyl chloride into ethylene and hydrochloric acid. So, the rate constant can be calculated for first order reaction.

Complete step by step answer:
Now, we are given with the thermal decomposition reaction of ethyl chloride; the reaction is

C$_2$H$_5$Cl$_g$ $\xrightarrow{h\nu}$ C$_2$H$_4$$_{(g)}$ + HCl $_{(g)}$

From the table, we can say that the value of initial pressure is 0.30 atm at the time 0 sec, and the value of final pressure is 0.50 atm at the time 300 sec.
As we know, the value of rate constant is represented by k in terms of pressure; i.e.

k = $\dfrac{2.303}{t}$log$\dfrac{P_0}{P_0 - P}$

 C$_2$H$_5$Cl$_g$ $\xrightarrow{h\nu}$ C$_2$H$_4$$_{(g)}$ + HCl $_{(g)}$



Now, from the above table we can calculate the total pressure at time t, i.e.
Total pressure, P$_t$ = (P$_0$ - p) + p + p
We have, P$_t$ = P$_0$ + p
We can calculate the value p, i.e. p = P$_t$ - P$_0$
Thus, value of pressure at time t for ethyl chloride;
P$_0$ - p = P$_0$ - P$_t$ + P$_0$
P$_0$ - p = 2 P$_0$ - P$_t$
Thus, now we will calculate the value of k for first order reaction,

k = $\dfrac{2.303}{t}$log$\dfrac{P_0}{2P_0 – P_t}$

Here, we have t = 300 sec, final pressure (P$_t$) = 0.5atm, and initial pressure (P$_0$) = 0.3atm.
If we substitute these value in the rate constant formula, then

k = $\dfrac{2.303}{300}$log$\dfrac{0.3}{2(0.3) – 0.5}$ ,

k = 3.6 $\times$ 10$^{-3}$ sec$^{-1}$

Therefore, in the end we can say that the value of k (rate constant) is 3.6 $\times$ 10$^{-3}$ sec$^{-1}$.

Note: Don’t get confused while finding the value of rate constant. Just draw the table to differentiate between the different values for decomposed molecules. With the help of a table; we are able to calculate the values of pressure, without any confusion.