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The decomposition of N_(2)O_(5) in CCl_(4) solution at 318 K 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 M and after 184 minutes, it is reduced to 2*08 M. The reaction takes place according tothe equation : 2N_(2)O_(5)to4NO_(2)+O_(2) Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO_(2) during this period ? |
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Answer» Solution :Average Rate `=-(1)/(2)(Delta[N_(2)O_(5)])/(Deltat)=-(1)/(2)((2*08-2*33)MOLL^(-1))/(1*84min)=6*79xx10^(-4)molL^(-1)min^(-1)` `=(6*79xx10^(-4)molL^(-1))/(min)xx(1min)/(60s)=1.13xx10^(-5)molL^(-1)s^(-1)` `=(6*79xx10^(-4)molL^(-1))/(min)xx(60min)/(1HR)=4*07xx10^(-2)molL^(-1)hr^(-1)` Rate `=(1)/(4)(Delta[NO_(2)])/(Deltat)=-(1)/(2)(Delta[N_(2)O_(5)])/(Deltat)=6*79xx10^(-4)molL^(-1)min^(-1)""("calculatedabove")` `:." Rate of production of "NO_(2),(Delta[NO_(2)])/(Deltat)=4xx6*79xx10^(-4)molL^(-1)min^(-1)=2*72xx10^(-3)molL^(-1)min^(-1)` |
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