1.

The decomposition of N_(2)O_(5) in C Cl_(4) at 318 K is studied by monitoring the concentration of N_(2)O_(5) in the solution. Initially, the concentration of N_(2)O_(5) is 2.4" mol L"^(-1) and after 200 minutes, it is reduced to 2.00" mol L"^(-1). What is the rate of production of NO_(2) during this period in mol L"^(-1)"min"^(-1) ?

Answer»

`4xx10^(-3)`
`2xx10^(-3)`
`1xx10^(-3)`
`2xx10^(-4)`

SOLUTION :`2N_(2)O_(5) to NO_2) + O_(2)`
Rate of reaction `= -(1)/(2) (d[N_(2)O_(5)])/(dt) = + (1)/(4)(d[NO_(2)])/(dt)"" …(i)`
Rate of DISAPPERANCE of `N_(2)O_(5) ((-d[N_(2)O_(5)])/(dt))`
` = ((2*4 - 2*0))/(20 min) mol L^(-1)`
`=(0*4)/(200) mol L^(-1) min^(-1)`
` = 2 xx 10^(-3) mol L^(-1) min^(-1)`
From EQU. (i),
Rate of production of `NO_(2)(+(d[NO_(2)])/(dt))`
`= -4 xx(1)/(2)(d[N_(2)O_(5)])/(dt)`
`=-2 xx(-2 xx 10^(-3)) mol L^(-1) min^(-1)`
`=4 xx 10^(-3) mol L^(-1) min^(-1)`


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