Question

The following data in the table were obtained during the first order thermal decomposition of $S{O_2}C{l_2}$ at a constant volume:
$S{O_2}C{l_2}(g) \to S{O_2}(g) + C{l_2}(g)$
Calculate the rate constant. Given: $\log 4 = 0.6021,\log 2 = 0.3010$

Experiment	Time/$s^{-1}$	Total pressure/atm
1	0	0.4
2	100	0.7

Answer 1

Hint: As we all know that the first order reactions are those which depend on the concentration of only one reactant and hence it is also called unimolecular reactions. The rate constant of a reaction can be precisely defined as the rate of a reaction when the concentration of the reactant becomes unity.

Complete step by step answer:
First we should understand the meaning of an order of a reaction which is basically the powers raised in different concentration terms in a rate expression with respect to the reactant. We should also know the fact that only that reactant whose concentration changes with passage of time are involved in rate expression therefore, if any reactant is present in excess then the order of reaction with respect to that reactant will be zero order.
Now, the first order reactions are the one that depends on the concentration of only one reactant and we should also understand some important cases of first order reaction. So, let us consider the given reaction and understand the conditions related:
$S{O_2}C{l_2}(g) \to S{O_2}(g) + C{l_2}(g)$.

	$S{O_2}C{l_2}(g)$	$S{O_2}(g)$	$C{l_2}(g)$
At time t = 0	${P^o}$	0	0
At time t = t	${P^o} - p$	P	P

After t time, the total pressure (Pt) will be given as:
${P_t} = ({P^o} - p) + p + p \\
{P_t} = {P^o} + p$
On rearranging, we get:
$p = {P_t} - {P^o} \\
{\text{ }} = {P^o} - ({P_t} - {P^o}) \\
{\text{ = 2}}{{\text{P}}^o} - {P_t}$
And the rate constant for this first order reaction will be given as:
$k = \dfrac{{2.303}}{t}\log \dfrac{{{P^o}}}{{2{P^o} - {P_t}}}$
Now, substituting the given values, $t = 100$, ${P^o} = 0.4$ and ${P_t} = 0.7$
$k = \dfrac{{2.303}}{t}\log \dfrac{{{P^o}}}{{2{P^o} - {P_t}}} \\
= \dfrac{{2.303}}{{100}}\log \dfrac{{0.4}}{{0.8 - 0.7}} \\
= \dfrac{{2.303}}{{100}}\log \dfrac{{0.4}}{{0.1}} \\
= 1{\text{.39}} \times {\text{1}}{{\text{0}}^{ - 2}}{s^{ - 1}}$

Therefore, the rate constant is found to be $1.39 \times {10^{ - 2}}{s^{ - 1}}$.

Note: If volume of reagent is given which reacts only with reactant at a given time interval then the rate constant can be calculated as $k = \dfrac{{2.303}}{t}\log \dfrac{{{V_1}}}{{{V_2}}}$. When temperature is increased, kinetic energy of molecules increases which activates the number of molecules and in turn the rate of reaction is increased ultimately increasing the rate constant.