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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)

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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 $_{(g)}$

At time t = 0 | P$_0$ | 0 | 0 |

At time t= t | P$_0$ - p | p | p |

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$^{-1}$