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The decomposition of N_(2)O_(5) in CCl_(4)at 318K has been studied by monitoring the concentration of N_(2)O_(5) in the solution.Initially the concentration of N_(2)O_(5) is 2.33 mol L^(1) and after 184 minutes ,it is reduced of 2.08 mol L^(-1) .The reaction take place according to the equation 2N_(2)O_(5(g))to4NO_(2(g))+O_(2(g)) Calculate the average rate of this reaction in terms of hours,minutes and seconds.What is the rate of production of N_(0)2 during this period? |
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Answer» Solution :The balance equation of reactants are as under: `2N_(2)O_(5(g))to4NO_(2(g))+O_(2(g))` Average Rate=`-(1)/(2){(Delta[N_(2)O_(5)])/(DELTAT)}` `(1)/(4)((Delta[NO_(2)])/(Deltat))` `(Delta[O_(2)])/(Deltat)` Initial t=0,the concentration of `N_(2)O_(5)=2.33 ml L^(-1)` after 184 minutes ,the concentration of `N_(2)O_(5)=2.08 MOL L^(-1)` So, `Deltat`=(184-0.0) minute=184 minute `Delta[R]=[R_(2)]-[R_(1)]` `therefore` Avereage rate =`-(1)/(2)(Delta[N_(2)O_(5)])/(Deltat)` `-(1)/(2)(-0.25 mol L^(1))/(184 min)` `(-0.25)/(2xx(184xx60)s)` `=1.1322xx10^(-5) mol L^(-1)S^(-1)` `r_(av)`(in HOUR)=`-(1)/(2) (Delta[R])/(Deltat ("in hour"))` `=-(-0.25 mol L^(-1))/(2xx(184+60)h)` `(+0.25xx60)/(2xx184) mol L^(-1) h^(-1)` |
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