For the order thermal decomposition reaction, the following data were obtained : C_(2)H_(5)Cl(g)rarrC_(2)H_(4)(g)+HCl (g){:("Time/Sec","Total pressure /atm"),(0,""0.30),(300,""0.50):} Calculate the rate constant. (Given: log2 = 0.301, log3 = 0.4771 and log4 = 0.6021)

For the order thermal decomposition reaction, the following data were

1.	For the order thermal decomposition reaction, the following data were obtained : C_(2)H_(5)Cl(g)rarrC_(2)H_(4)(g)+HCl (g){:("Time/Sec","Total pressure /atm"),(0,""0.30),(300,""0.50):} Calculate the rate constant. (Given: log2 = 0.301, log3 = 0.4771 and log4 = 0.6021)
Answer» <P> Solution :Suppose , the initial pressure of the reactant is P and its pressure reduces to x at a later time t , after the REACTION has started . So, at time t , the partial pressures of different components in the reaction mixture will be as follows: `""C_(2)H_(5)Cl(g)rarrC_(2)C_(4)(g)+HCl(g)` `{:("At " t = 0,P,,),("At "t = t,P-x,""x,""x):}` Total pressure of the mixture at t, `P_(t)=P-x+x+x=P+x` This gives`x=P_(t)-P` Therefore, at time t, the partial pressure of the reactant, `P_(i)=P-x=P-P_(t)+P=2P-P_(t)` As the reaction follows first order kinetics , its integrated rate law is - `k=(2.303)/(t)log.([A]_(0))/([A])` Substituting partial pressures for concentrations , we have, `k=(2.303)/(t)log.(P)/(P_(i))=(2.303)/(t)log.(P)/(2P-P_(t))` Given that `P = 0.30 "atm" and P_(t) = 0.50 "atm at " t = 300s` Substituting these values into the equation of k , we GET, `k=(2.303)/(300)log.(0.3)/(2xx0.3-0.5)=3.662xx10^(-3)s^(-1)`