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InterviewSolution
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The decompoistion of `N_(2)O_(5)` in `C CI_(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` is `2.33 M` and after `184 min`, it is reduced to `2.08 M`. The reaction takes place according to the equation: `2N_(2)O_(5) rarr 4NO_(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» Average reaction rate `= -(1)/(2) ((Delta[N_(2)O_(5)])/(Delta t))` `= -(1)/(2) (([2.08-2.33]mol L^(-1))/(184 min))` `= -(1)/(2) ((-0.25 mol L^(-1))/(184 min))` `= 6.79xx10^(-4)mol L^(-1) min^(-1)` Average reaction rate in seconds `= 6.79xx10^(-4) (mol L^(-1))/(min)xx(1min)/(60s)` `= 0.113xx10^(-4) mol L^(-1)s^(-1)` `= 1.13xx10^(-5)mol L^(-1)s^(-1)` Average reaction rate in hours `= 6.79xx10^(-4)(mol L^(-1))/(min)xx(60min)/(1h)` `= 407.4xx10^(-4)mol L^(-1)h^(-1)` `= 4.07xx10^(-2) mol L(-1)h^(-1)` |
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