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InterviewSolution
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The decomposition of `N_(2)O_(5)` in `C Cl_(4)` at 318K has been studies 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 to `2.08 mol L^(-1).` The reaction takes placed according to the equation `2N_(2)O_(5)(g)to 4 NO _(2)(g) +O_(2)(g)` Calculate the average rate of this reaction in terms of hours, minutres and seconds. What is the rate of proudction of `NO_(2)` during this period ? |
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Answer» Average Rate `=1/2 {-(Delta[N_(2)O_(5)])/(Deltat)}=-1/2[((0.28-2.33)molL^(-1))/(184 min)]` `=6.79 xx10^(-4) mol L^(-1) //min` `=(6.79xx10^(-4) mol L^(-1)min ^(-1))xx(60 min //1h)` `=4.07 xx10^(-2) mol L^(-1) //h` `=6.79 xx10^(-4)mol L^(-1)xx 1min//60s =1.13 xx10^(-5)mol L^(-1)s^(-1)` It may be remembered that Rate `=1/4 {(Delta[NO_(2)])/(Deltat)}` `(Delta[NO_(2)])/(Deltat)=6.79xx10^(-4)xx4 mol L^(-1) min ^(-1)` `=2.72xx10^(-3)mol L^(-1) min^(-1)` |
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