Usethe data provided to graphically determine the order and rate constant of the following reaction: \({\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}} \to {\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{ + C}}{{\bf{l}}_{\bf{2}}}\)

Time(hr)	0	5.00*\({\bf{1}}{{\bf{0}}^{\bf{3}}}\)	1.00*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)	1.50*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)	2.50*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)	3.00*104	4.00*104
\({\bf{(S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{)}}\)(M)	0.100	0.0896	0.0802	0.0719	0.0577	0.0517	0.0415

In first-order kinetics, the rate of reaction is directly proportional to the concentration of reactant.

\({\bf{Rate of reaction}} \propto \left( {{\bf{ concentration of reactant}}} \right)\)

\({\bf{Rate of reaction = K }}\left( {{\bf{concentration of reactant}}} \right)\)

Where K is the rate constant.

The rate constant of a first-order reaction is given as

\(k{\bf{ }} = \frac{{\ln \frac{{{{(P)}_0}}}{{(P)}}}}{t}\)

Now, the slope of the plot is given as

\(\begin{align}Slope &= \frac{{( - 2.412) - ( - 2.302)}}{{(5 \times {{10}^3}) - 0}}\\ &= - 2.20 \times {10^{ - 5}}{s^{ - 1}}\end{align}\)

The slope of the curve for the variation in the concentration vs time plot is equal to the negative of the rate constant or the reaction.

So,

\(\begin{align}Rate &= - \left( { - 2.20 \times {{10}^{ - 5}}{s^{ - 1}}} \right)\\ &= 2.20 \times {10^{ - 5}}{s^{ - 1}}\end{align}\)

The plot of ln(P) vs time is a linear plot with a negative slope. This indicates first-order reaction kinetics.

Thus, the reaction is first order, and the rate constant is \({\bf{2}}{\bf{.2 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)