Question

The following data were obtained during the first order thermal decomposition of \[S{O_2}C{l_2}\] at a constant volume.
$S{O_2}C{l_2}(g) \to S{O_2}(g) + C{l_2}(g)$

Experiment	Time/s	Total pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of the reaction when total pressure is\[0.65atm\].

Answer 1

Hint: The order of reaction refers to the power dependence of rate of the reaction with the concentration of reactants. For first order reaction, the rate of reaction is dependent on the concentration of the single reactant.

Complete step by step answer:
As the first order reaction depends on the concentration of only one species, the rate constant of the reaction does not depend on the concentration of reactants.
For the given experiment of decomposition of\[S{O_2}C{l_2}\], the initial pressure at time \[t = 0s\] is \[0.5atm\]. The total pressure at time \[t = 100s\] is \[0.6atm\].
Let us consider the initial pressure is \[{P_0}\] and pressure at time t is\[{P_t}\] . Thus for the reaction
$S{O_2}C{l_2}(g) \to S{O_2}(g) + C{l_2}(g)$

	\[S{O_2}C{l_2}\left( g \right)\]	\[S{O_2}\left( g \right)\]	\[C{l_2}\left( g \right)\]
At t = 0	\[{P_0}\]	\[0\]	\[0\]
At t = t	\[{P_0} - p\]	\[p\]	\[p\]

The rate of reaction in terms of pressure is
\[k = \dfrac{{2.303}}{t}\log \dfrac{{{P_0}}}{{{P_0} - p}}\]
The total pressure can be evaluated as the sum of the individual pressure of \[S{O_2}C{l_2}\left( g \right)\], \[S{O_2}\left( g \right)\] and \[C{l_2}\left( g \right)\].
Thus pressure at time t, ${P_t} = ({P_0} - p) + p + p = {P_0} + p$
Or, $p = {P_t} - {P_0}$
Therefore, ${P_0} - p = {P_0} - ({P_t} - {P_0}) = 2{P_0} - {P_t}$
The rate of the reaction becomes, \[k = \dfrac{{2.303}}{t}\log \dfrac{{{P_0}}}{{2{P_0} - {P_t}}}\]
Inserting the values of \[t\], \[{P_0}\] and \[{P_t}\] in the first order rate equation,
$k = \dfrac{{2.303}}{{100}}\log \dfrac{{0.5}}{{(2 \times 0.5) - 0.6}}$
$k = 2.23 \times {10^{ - 3}}{s^{ - 1}}$ .
Now when the total pressure is \[0.65atm\],
\[{P_{S{O_2}C{l_2}}} = 2{P_0} - Pt = (2 \times 0.5) - 0.65 = 0.35atm\]
Rate =$k \times {P_{S{O_2}C{l_2}}} = 2.23 \times {10^{ - 3}} \times 0.35 = 7.8 \times {10^{ - 4}}atm{s^{ - 1}}$ .
Hence the rate of the reaction when total pressure is \[0.65atm\] is $7.8 \times {10^{ - 4}}atm{s^{ - 1}}$.

Note: A unique characteristic of first order reaction is that the half life is independent of the initial concentration of the reactant. Also the rate constant for the reaction is inversely proportional to the half life of the reaction.