The decomposition of `N_(2) O_(5)` in `"CCI"_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 2.08M. The reaction takes place according to the equation, `2N_(2) O_(5) to 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.

Explanation

Average rate `= - (1)/(2) (Delta [N_(2) O_(5) ] )/( Delta t)`
`= - (1)/(2) ((2.08 - 2.33))/(1.84" min" )" mol L"^(-1)`
`= 6.79 xx 10^(-4)" mol L"^(-1)" min"^(-1)`
`= (6.79 xx 10^(-4)" mol L"^(-1) )/("min") xx (1" min")/(60" s")`
`= 1.13 xx 10^(-5)" mol L"^(-1)" s"^(-1)`
Similarly, average rate `= (6.79 xx 10^(-4)" mol L"^(-1) )/(" min" ) xx (60" min")/(1" hr" )`
`= 13 xx 10^(-5)" mol L"^(-1)" s"^(-1)`
Rate `= (1)/(4) (Delta [NO_2])/(Delta t) = (-1)/(2) (Delta[N_(2) O_(5) ])/(Delta t)`
`=6.79 xx 10^(-4)" mol L"^(-1)" min"^(-1)`
`therefore` Rate of production of `NO_2`,
`(Delta [NO_2])/(Delta t) = 4 xx 6.79 xx 10^(-4)" mol L"^(-1)" min"^(-1)`
`=2.72 xx 10^(-5)" mol L"^(-1)" min"^(-1)`