Question

The following data were obtained during the first order thermal decomposition of $S{{O}_{2}}C{{l}_{2\left( g \right)}}$​ at a constant volume.
$S{{O}_{2}}C{{l}_{2\left( g \right)}}\to S{{O}_{2\left( g \right)}}+C{{l}_{2\left( g \right)}}$

Experiment	Time / ${{s}^{-1}}$	Total pressure / atm
$1$	$0$	$0.5$
$2$	$100$	$0.6$

Answer 1

We are given with the values of pressure, at the different t 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 solution:
Now, we are given with the thermal decomposition reaction of $S{{O}_{2}}C{{l}_{2\left( g \right)}}$
$S{{O}_{2}}C{{l}_{2\left( g \right)}}\to S{{O}_{2\left( g \right)}}+C{{l}_{2\left( g \right)}}$
As we know, the value of rate constant is represented by k in terms of pressure; i.e.
$k=\dfrac{2.303}{t}\log \dfrac{{{P}_{o}}}{{{P}_{o}}-{{P}_{t}}}$
$S{{O}_{2}}C{{l}_{2\left( g \right)}}\to S{{O}_{2\left( g \right)}}+C{{l}_{2\left( g \right)}}$

| $S{{O}_{2}}C{{l}_{2\left( g \right)}}$| $S{{O}_{2\left( g \right)}}$| $C{{l}_{2\left( g \right)}}$
---|---|---|---
At time $t=0$| ${{P}_{o}}$| $0$| $0$
At time $t=t$| ${{P}_{o}}-p$| $p$| $p$

Now, from the above table we can calculate the total pressure at time t, i.e.
Total pressure, ${{P}_{t}}=\left( {{P}_{o}}-p \right)+p+p$
We have, ${{P}_{\text{t}}}={{P}_{o}}+p$
We can calculate the value p, i.e. $p={{P}_{\text{t}}}-{{P}_{o}}$
Thus, value of pressure at time t for $S{{O}_{2}}C{{l}_{2\left( g \right)}}$
${{P}_{o}}-p={{P}_{o}}-{{P}_{\text{t}}}+{{P}_{o}}$
$\Rightarrow {{P}_{o}}-p=2{{P}_{o}}-{{P}_{\text{t}}}$
Thus, now we will calculate the value of k for
$k=\dfrac{2.303}{t}\log \dfrac{{{P}_{o}}}{2{{P}_{o}}-{{P}_{t}}}$
Here we have $Rate~=k{{P}_{S{{O}_{2}}C{{l}_{2}}}}$ and thus the total pressure is $0.65atm$
${{P}_{S{{O}_{2}}C{{l}_{2}}}}=2{{p}_{o}}-{{p}_{t}}=2\times 0.5-0.65=0.35atm$
Hence, $rate=2.23\times {{10}^{-3}}\times 0.035=7.805\times {{10}^{-4}}$
$\Rightarrow rate=7.805\times {{10}^{-4}}atm/s$

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